Differential equations — Paper 2 · A Level Mathematics - Further 9231

1 Find the solution of the differential equation d y -7 x + 5y = e d x for which y = 0 when x = 0 . Give your answer in the form y = f ( x) . [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 1 5d 5 e e =  x x M1 A1 ( ) 5 2 d e e d − = x x y x M1 5 2 1 2 e e− = − + x x y C A1 1 2 0 = −+ C M1 5 7 1 1 2 2 e e − − = − x x y A1 6

This question in 9231/21 May/June 2020

7 It is given that x = t 3 y and 3 d 2 y 3 2 d y 3 2 2 + 4t + 6t + 13t + 12t + 6 t y = 61e . t 2 1 t d t ` j d t ` j (a) Show that d 2 x d x 21 t + 4 + 13x = 61e . [4] 2 d t d t … … … … … … … … … … … … … … … … … … … … … … (b) Find the general solution for y in terms of t. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 3 2 d d 3 d d = + x y t t y t t B1 2 2 3 2 2 2 d d d 6 6 d d d = + + x y y t t ty t t t B1 2 2 3 2 3 2 3 2 2 d d d d d 4 13 6 6 4 12 13 d d d d d + + = + + + + + x x y y y x t t ty t t y t y t t t t t 1 2 61e = t M1 A1 4 7(b) 2 4 13 0 2 3 + + =  = −± m m m i M1 ( ) 2 e cos3 sin3 − = + t x A t B t A1 1 1 1 2 2 2 1 1 2 4 e e e =  =  =   t t t x k x k x k B1 1 4 2 13 61 4 + + =  = k k k k M1 A1 ( ) ( ) 1 1 2 2 3 2 3 2 3 e cos3 sin3 4e e cos3 sin3 4 e − − − − = + +  = + + t t t t t y A t B t y t A t B t t M1 A1 7

This question in 9231/21 May/June 2020

1 Find the solution of the differential equation d y -7 x + 5y = e d x for which y = 0 when x = 0 . Give your answer in the form y = f ( x) . [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 1 5d 5 e e =  x x M1 A1 ( ) 5 2 d e e d − = x x y x M1 5 2 1 2 e e− = − + x x y C A1 1 2 0 = −+ C M1 5 7 1 1 2 2 e e − − = − x x y A1 6

This question in 9231/22 May/June 2020

7 It is given that x = t 3 y and 3 d 2 y 3 2 d y 3 2 2 + 4t + 6t + 13t + 12t + 6 t y = 61e . t 2 1 t d t ` j d t ` j (a) Show that d 2 x d x 21 t + 4 + 13x = 61e . [4] 2 d t d t … … … … … … … … … … … … … … … … … … … … … … (b) Find the general solution for y in terms of t. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 3 2 d d 3 d d = + x y t t y t t B1 2 2 3 2 2 2 d d d 6 6 d d d = + + x y y t t ty t t t B1 2 2 3 2 3 2 3 2 2 d d d d d 4 13 6 6 4 12 13 d d d d d + + = + + + + + x x y y y x t t ty t t y t y t t t t t 1 2 61e = t M1 A1 4 7(b) 2 4 13 0 2 3 + + =  = −± m m m i M1 ( ) 2 e cos3 sin3 − = + t x A t B t A1 1 1 1 2 2 2 1 1 2 4 e e e =  =  =   t t t x k x k x k B1 1 4 2 13 61 4 + + =  = k k k k M1 A1 ( ) ( ) 1 1 2 2 3 2 3 2 3 e cos3 sin3 4e e cos3 sin3 4 e − − − − = + +  = + + t t t t t y A t B t y t A t B t t M1 A1 7

This question in 9231/22 May/June 2020

1 Find the general solution of the differential equation d 2 x d x 8 t - 8 - 9x = 9e . [6] 2 d t d t … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 1 9 0 1,9 m m m − −=  =− M1 9 e e t t x A B − = + A1 8 8 8 8 64 t t t x ke x ke x ke =  =  =   B1 8 8 8 8 64 e 64 e 9 e 9e 9 9 t t t t k k k k − − = − = M1 1 k = − A1 9 8 e e e t t t x A B − = + − A1 6

This question in 9231/23 May/June 2020

7 (a) Show that an appropriate integrating factor for 2 d y 2 2 2 ( x + 1 ) + y x + 1 = x - x x + 1 d x is x + x 2 + 1 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the solution of the differential equation 2 d y 2 2 2 ( x + 1 ) + y x + 1 = x - x x + 1 d x for which y = ln 2 when x = 0 . Give your answer in the form y = f ( x) . [7] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) ( ) 2 2 2 d 1) d 1 1) y y x x x x x x + = −√( + + √( + B1 1 2 1 d sinh 1) e e x x x − √( +  = M1 A1 2 1) x x = +√( + A1 4 7(b) ( ) ( ) 2 d 1) d y x x x + √( + = 2 1 x x − + M1 A1 ( ) ( ) 2 2 1 2 2 1) d ln 1 1 x y x x x x C x + √( + = − = − + + +  M1 A1 ln 2 C = M1 ( ) ( ) ( ) 2 1 2 2 2 1 2 2 ln2 ln 1 1) ln 1) 1) x y x x x x+ x − + = = −√( + √( + √( + M1 A1 7

This question in 9231/23 May/June 2020

2 The variables x and y are related by the differential equation d 2 y dy 2 9 2 + 6 + y = 3x + 30x. dx dx (a) Find the general solution for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … (b) State an approximate solution for large positive values of x. [1] … … … …

7 marks

Mark scheme: 2(a) 2 1 9 6 1 0 3 + + =  = − m m m M1 Auxiliary equation. ( ) 1 3 e − = + x y Ax B A1 Complimentary function. 2 ' 2 '' 2 = + +  = +  = y p qx rx y q rx y r B1 Particular integral and its derivatives. 2 2 18 6 12 3 30 + + + + + = + r q rx p qx rx x x M1 Substitutes and equates coefficients. 3, 6, 18 = = − = − r q p A1 ( ) 2 1 3 e 3 6 18 − = + + − − x y Ax B x x A1 6 2(b) 2 3 6 18 = − − y x x B1 FT 1

This question in 9231/21 Oct/Nov 2020

i - 4 cos 2i + 3) . [5]6 (a) Use de Moivre’s theorem to show that sin 4 i = 18 ( cos 4 … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the solution of the differential equation d y 3 + y cot i = sin i di r . [6] for which y = 0 when i = 12 … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6(a) 1 2isinθ − − = z z B1 Use of 1 2isinθ − − = z z . ( ) ( ) ( ) 4 1 4 4 2 2 4 6 − − − − = + − + + z z z z z z M1 A1 Expands and groups. ( ) ( ) ( ) 4 2isin 2cos4 4 2cos2 6 θ θ θ = − + M1 Substitutes 2cos θ − + = n n z z n ( ) 4 1 cos4 4cos2 3 sin 8 θ θ θ = − + A1 AG 5 6(b) cot lnsin sin θ θ θ θ = =  d e e M1 A1 Finds integrating factor. ( ) ( ) 4 1 cos4 4cos2 3 d sin sin 8 d θ θ θ θ θ = − = + y M1 Correct form on LHS and uses identity given in (a). 1 1 sin 4 2sin 2 3 8 4 sinθ θ θ θ   = − + +     C y A1 0 π 1 3 8 2 C   = +     M1 Substitutes initial conditions. 1 1 3 sin 4 2sin 2 3 in π s 8 4 2 y θ θ θ θ   = − + −     A1 OE 6

This question in 9231/21 Oct/Nov 2020

4 Find the solution of the differential equation dy x + 2y = ex dx for which y = 3 when x = 1. Give your answer in the form y = f ( x) . [8] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 d 2 1 e d + = x y y x x x B1 Divides through by x. 1 2 2 d −  = x x e x M1 A1 Finds integrating factor, must be integrating 1. − x ( ) 2 d e d = x yx x x M1 Correct form on LHS and attempt to integrate 1 ex x multiplied by their integrating factor. 2 ( 1)e = − + x yx x C A1 3 = C M1 Finds C. 2 ( 1)e 3 − + = x x y x M1 A1 Divides through by coefficient of y. 8

This question in 9231/22 Oct/Nov 2020

6 Find the particular solution of the differential equation d 2 x dx + 8 + 15 x = 102 cos 3 t , 2 dt dt dx given that, when t = 0 , x = 1 and = 0 . [11] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6 2 8 15 0 3, 5 + + =  = −− m m m M1 Auxiliary equation. 5 3 e e − − = + t t x A B A1 Complementary function. sin3 cos3 3 cos3 3 sin3 9 sin3 9 cos3 = +  = −  = − −   x p t q t x p t q t x p t q t M1 A1 Particular integral and its derivatives. 9 24 15 0 6 24 0 − − + =  − = p q p p q 9 24 15 102 4 17 − + + =  + = q p q p q M1 Substitutes and equates coefficients. 4, 1 = = p q A1 5 3 e e 4sin3 cos3 − − = + + + t t x A B t t A1 5 3 5 e 3 e 12cos3 3sin3 − − = − − + −  t t x A B t t M1 Differentiating the correct form. 1 1 = + + A B 0 5 3 12 6, 6 = − − +  = = − A B A B M1 A1 Forms simultaneous equations using initial conditions. 5 3 6e 6e 4sin3 cos3 − − = − + + t t x t t A1 11

This question in 9231/22 Oct/Nov 2020

2 The variables x and y are related by the differential equation d 2 y dy 2 9 2 + 6 + y = 3x + 30x. dx dx (a) Find the general solution for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … (b) State an approximate solution for large positive values of x. [1] … … … …

7 marks

Mark scheme: 2(a) 2 1 9 6 1 0 3 + + =  = − m m m ( ) 1 3 e − = + x y Ax B A1 Complimentary function. 2 ' 2 '' 2 = + +  = +  = y p qx rx y q rx y r B1 Particular integral and its derivatives. 2 2 18 6 12 3 30 + + + + + = + r q rx p qx rx x x M1 Substitutes and equates coefficients. 3, 6, 18 = = − = − r q p A1 ( ) 2 1 3 e 3 6 18 − = + + − − x y Ax B x x A1 6 2(b) 2 3 6 18 = − − y x x B1 FT 1

This question in 9231/23 Oct/Nov 2020

i - 4 cos 2i + 3) . [5]6 (a) Use de Moivre’s theorem to show that sin 4 i = 18 ( cos 4 … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the solution of the differential equation d y 3 + y cot i = sin i di r . [6] for which y = 0 when i = 12 … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6(a) 1 2isinθ − − = z z B1 Use of 1 2isinθ − − = z z . ( ) ( ) ( ) 4 1 4 4 2 2 4 6 − − − − = + − + + z z z z z z M1 A1 Expands and groups. ( ) ( ) ( ) 4 2isin 2cos4 4 2cos2 6 θ θ θ = − + M1 Substitutes 2cos θ − + = n n z z n ( ) 4 1 cos4 4cos2 3 sin 8 θ θ θ = − + A1 AG 5 6(b) cot lnsin sin θ θ θ θ = =  d e e M1 A1 Finds integrating factor. ( ) ( ) 4 1 cos4 4cos2 3 d sin sin 8 d θ θ θ θ θ = − = + y M1 Correct form on LHS and uses identity given in (a). 1 1 sin 4 2sin 2 3 8 4 sinθ θ θ θ   = − + +     C y A1 0 π 1 3 8 2 C   = +     M1 Substitutes initial conditions. 1 1 3 sin 4 2sin 2 3 in π s 8 4 2 y θ θ θ θ   = − + −     A1 OE 6

This question in 9231/23 Oct/Nov 2020

2 The variables x and y are related by the differential equation d 2 y dy + 3 + 2 y = 2 x + 1. 2 dx dx (a) Find the general solution for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … (b) State an approximate solution for large positive values of x. [1] … … …

7 marks

Mark scheme: 2(a) 2 3 2 0 m m + + = leading to m = -1, -2 M1 Auxiliary equation. 2 e e x x y A B − − = + A1 Complimentary function. leading to ' leading to '' 0 y p qx y q y = + = = B1 Particular integral and its derivatives. 3 2( ) 2 1 q p qx x + + = + M1 Substitutes and equates coefficients. 1 1 q p = = − A1 2 e e 1 x x y A B x − − = + + − A1 Must see ‘y =’. 6 Question Answer Marks Guidance 2(b) 1 = − y x B1 FT Accept 1 ≈ − y x . 1

This question in 9231/21 May/June 2021

4 Find the solution of the differential equation d y 1 sin i i , + y = tan 2 di r . Give your answer in the form y = f( i) . [9] where 0 1 i 1 r , given that y = 1 when i = 12 [You may use without proof the result that 1 i .] cosec i d i = lntan 2 y … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4 1 2 d cosec tan sin d θ θ θ θ = + y y θ 1 2 ln tan 1 osec d 2 c e e tan θ θ θ θ = =  M1 A1 Finds integrating factor. ( ) 1 1 1 2 2 2 1 1 1 2 2 2 2 1 2 1 2 3 2 s d tan sin tan ec sin cos tan d y θ θ θ θ θ θ θ θ   = =   =   M1 Correct form on LHS and uses an appropriate identity. 2 1 1 1 2 2 2 tan tan y C θ θ = + ( ) 1 1 2 2 2 or sec C θ + M1 A1 Integrates RHS. 1 1 2 C = + M1 Substitutes initial conditions. ( )[ ] 2 1 1 2 2 1 tan cose cot c y θ θ θ = + = M1 A1 Divides through by their integrating factor. 9

This question in 9231/21 May/June 2021

2 The variables x and y are related by the differential equation d 2 y dy + 3 + 2 y = 2 x + 1. 2 dx dx (a) Find the general solution for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … (b) State an approximate solution for large positive values of x. [1] … … …

7 marks

Mark scheme: 2(a) 2 3 2 0 m m + + = leading to m = -1, -2 M1 Auxiliary equation. 2 e e x x y A B − − = + A1 Complimentary function. leading to ' leading to '' 0 y p qx y q y = + = = B1 Particular integral and its derivatives. 3 2( ) 2 1 q p qx x + + = + M1 Substitutes and equates coefficients. 1 1 q p = = − A1 2 e e 1 x x y A B x − − = + + − A1 Must see ‘y =’. 6 Question Answer Marks Guidance 2(b) 1 = − y x B1 FT Accept 1 ≈ − y x . 1

This question in 9231/22 May/June 2021

4 Find the solution of the differential equation d y 1 sin i i , + y = tan 2 di r . Give your answer in the form y = f( i) . [9] where 0 1 i 1 r , given that y = 1 when i = 12 [You may use without proof the result that 1 i .] cosec i d i = lntan 2 y … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4 1 2 d cosec tan sin d θ θ θ θ = + y y θ 1 2 ln tan 1 osec d 2 c e e tan θ θ θ θ = =  M1 A1 Finds integrating factor. ( ) 1 1 1 2 2 2 1 1 1 2 2 2 2 1 2 1 2 3 2 s d tan sin tan ec sin cos tan d y θ θ θ θ θ θ θ θ   = =   =   M1 Correct form on LHS and uses an appropriate identity. 2 1 1 1 2 2 2 tan tan y C θ θ = + ( ) 1 1 2 2 2 or sec C θ + M1 A1 Integrates RHS. 1 1 2 C = + M1 Substitutes initial conditions. ( )[ ] 2 1 1 2 2 1 tan cose cot c y θ θ θ = + = M1 A1 Divides through by their integrating factor. 9

This question in 9231/22 May/June 2021

5 The variables x and y are related by the differential equation d 2 y dy -x - 2 - 3y = 4 e . 2 dx dx (a) Find the value of the constant k such that y = kxe -x is a particular integral of the differential equation. [4] … … … … … … … … … … … … … … dy 1 (b) Find the solution of the differential equation for which y = = 2 when x = 0 . [6] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a) ( ) ( ) e leading to ' e 1 leading to '' e 2 x x x y kx y k x y k x − − − = = − = − B1 B1 Differentiates particular integral. ( ) ( ) e 2 2 e 1 3 e 4e leading to 4 4 x x x x k x k x kx k − − − − − − − − = − = M1 Substitutes. 1 = − k A1 4 5(b) 2 2 3 0 leading to 1,3 m m m − − = = − M1 Auxiliary equation. 3 e e − = + x x y A B A1 Complementary function. (Can be awarded if seen in the general solution.) 3 3 e e e ( )e e − − − = + − = − + x x x x x y A B x A x B A1 FT General solution. FT on their k. Must have “ = y ”. 3 ' ( )e e 3 e − − = − − − + x x x y A x B B1 Derivative of general solution. 1 1 1 2 2 2 , 1 3 leading to 0, A B A B A B + = − −+ = = = M1 Substitutes initial conditions into general solution. 3 1 2 e e− = − x x y x A1 Must have “ = y ”. 6

This question in 9231/23 May/June 2021

6 (a) Starting from the definitions of sinh and cosh in terms of exponentials, prove that 2 sinh 2 x = cosh 2 x - 1. [3] … … … … … … … … … … … (b) Find the solution to the differential equation d y + y coth x = 4 sinh x d x for which y = 1 when x = ln 3 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) ( ) ( ) 1 1 2 2 sinh e e cosh e e − − = − = + x x x x x x B1 Writes in exponential from. ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 2 e e e 2 e e e 1 − − − − = − + = + − x x x x x x M1 Expands. cosh 2 1 − x A1 AG. 3 6(b) lns oth inh c e e sinh = =  x x x M1 A1 Finds integrating factor. 2 d ( sinh ) 4sinh d = y x x x M1 Correct form on both sides. ( ) 1 2 sinh 2 sinh 2 sinh 2 2 = − + = − + y x x x C x x C M1 A1 Integrates 2 sinh x using identity. 4 40 sinhln3 sinhln9 2ln3 leading to 2ln3 3 9 C C = − + = − + M1 Substitutes initial conditions into their solution. 28 sinh sinh2 2 2ln3 9 = − + − y x x x A1 7

This question in 9231/23 May/June 2021

5 Find the particular solution of the differential equation d 2 y dy - 2 + y = 4 cos x, 2 dx dx dy given that, when x = 0 , y =- 4 and = 3 . [11] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5 2 2 1 0 1 − + =  = m m m ( ) e = + x y Ax B A1 Complementary function. Accept ‘y =’ missing. sin cos y p x q x = + , ' cos sin y p x q x = − , '' sin cos y p x q x = − − M1 A1 Particular integral and its derivatives. 2 0 − + + = p q p 2 4 −− + = q p q M1 Substitutes and equates coefficients. 2 = − p 0 = q A1 ( ) e 2sin = + − x y Ax B x A1 ( ) ' e e 2cos = + + − x x y A Ax B x M1 Differentiates. 4 = − B 2 3 9 + − =  = A B A M1 A1 Forms simultaneous equations using initial conditions. ( ) e 9 4 2sin = − − x y x x A1 11

This question in 9231/21 Oct/Nov 2021

7 (a) Show that an appropriate integrating factor for 2 dy 2 2 x - 1 + y = x - x x - 1 dx is x + x 2 - 1 . [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the solution of the differential equation 2 dy 2 2 x - 1 + y = x - x x - 1 dx = f ( x). [7] for which y = 1 when x = 54 . Give your answer in the form y … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 2 2 2 2 d 1 1 d 1 1 − − + = − − y x x x y x x x B1 Divides through by 2 1. − x 1 2 1 d cosh 1) e e − √( − =  x x x M1 A1 Finds integrating factor. M1 for correct form 2 1 d 1) e , √( −  a x x where a is a non-zero constant. 2 1 x x + − A1 AG 4 7(b) ( ) ( ) 2 2 d 1 d 1 + − = − x y x x x x M1 A1 Correct form on LHS and simplifies RHS. ( ) 2 2 1 1 + − = −+ y x x x C M1 A1 Integrates RHS. For M1, needs to integrate non-zero multiple of 2 . 1 − x x 3 4 2 C = + leading to 5 4 C = M1 Substitutes initial conditions. 5 2 4 2 1 1 −+ = + − x y x x M1 A1 Divides through by coefficient of y. 7

This question in 9231/21 Oct/Nov 2021

2 Find the solution of the differential equation dy 4 x 3 y + = 6x 4 dx x + 5 for which y = 1 when x = 1. Give your answer in the form y = f ( x) . [7] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 2 ( ) 4 ln 5 4 e 5 + = + x x ( ) ( ) 4 5 d 5 6 30 d + = + y x x x x M1 Writes in correct form. 4 6 2 ( 5) 15 + = + + y x x x C A1 6 16 = + C M1 Substitutes initial conditions. 6 2 4 15 10 5 + − = + x x y x M1 A1 Division through by coefficient of y. 7

This question in 9231/22 Oct/Nov 2021

7 It is given that y = x 2 w and 2 2 d w d w 2 2 x 2 + 4x ( x + 1) + ( 5x + 8x + 2) w = 5x + 4 x + 2 . d x d x (a) Show that d 2 y d y 2 + 4 + 5y = 5x + 4x + 2 . [4] 2 d x d x … … … … … … … … … … … … … … … … … … … … … … … (b) Find the general solution for w in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 2 d d 2 d d = + y w x xw x x B1 2 3 d 1 d 2 d d = − w y y x x x x 2 2 2 2 2 d d d 4 2 d d d = + + y w w x x w x x x B1 2 2 2 3 4 2 2 d 1 d 4 d 6 d d d = − + w y y y x x x x x x 2 2 2 2 2 2 2 d d d d d 4 5 4 2 4 8 5 d d d d d + + = + + + + + y y w w w y x x w x xw x w x x x x x M1 Uses substitution to find w-x equation, AG. ( ) ( ) 2 2 2 2 2 2 d d 5 4 4 5 8 2 5 4 2 d d w w x x x x x w x x x x + + + + + = + + A1 AG 4 7(b) 2 4 5 0 m m + + = leading to 2 m i = −± M1 Auxiliary equation. ( ) 2 e cos sin − = + x y A x B x A1 Complimentary function. Allow “ = y ” missing. Don’t allow e.g. ( ) 2 e cos isin . − + x A x B x 2 y px qx r = + + , ' 2 y px q = + , '' 2 y p = B1 Particular integral and its derivatives. 2 2 2 8 4 5 5 5 5 (8 5 ) (2 4 5 ) 5 5 8 5 4 2 4 5 2 + + + + + = + + + + + = + = + + = p px q px qx r px p q x p q r p p q p q r M1 Substitutes and equates coefficients. 1 = p 4 5 = − q 16 25 = r A1 ( ) 2 2 2 16 4 5 25 e cos sin − = + + − + x x w A x B x x x M1 Substitutes for y and find w in terms of x. ( ) ( ) ( ) 2 1 2 5 5 4 4 cos sin 1 ( ) − − − = + + − + x w xe A x B x x x A1 7

This question in 9231/22 Oct/Nov 2021

5 Find the particular solution of the differential equation d 2 y dy - 2 + y = 4 cos x, 2 dx dx dy given that, when x = 0 , y =- 4 and = 3 . [11] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5 2 2 1 0 1 − + =  = m m m ( ) e = + x y Ax B A1 Complementary function. Accept ‘y =’ missing. sin cos y p x q x = + , ' cos sin y p x q x = − , '' sin cos y p x q x = − − M1 A1 Particular integral and its derivatives. 2 0 − + + = p q p 2 4 −− + = q p q M1 Substitutes and equates coefficients. 2 = − p 0 = q A1 ( ) e 2sin = + − x y Ax B x A1 ( ) ' e e 2cos = + + − x x y A Ax B x M1 Differentiates. 4 = − B 2 3 9 + − =  = A B A M1 A1 Forms simultaneous equations using initial conditions. ( ) e 9 4 2sin = − − x y x x A1 11

This question in 9231/23 Oct/Nov 2021

7 (a) Show that an appropriate integrating factor for 2 dy 2 2 x - 1 + y = x - x x - 1 dx is x + x 2 - 1 . [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the solution of the differential equation 2 dy 2 2 x - 1 + y = x - x x - 1 dx = f ( x). [7] for which y = 1 when x = 54 . Give your answer in the form y … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 2 2 2 2 d 1 1 d 1 1 − − + = − − y x x x y x x x B1 Divides through by 2 1. − x 1 2 1 d cosh 1) e e − √( − =  x x x M1 A1 Finds integrating factor. M1 for correct form 2 1 d 1) e , √( −  a x x where a is a non-zero constant. 2 1 x x + − A1 AG 4 7(b) ( ) ( ) 2 2 d 1 d 1 + − = − x y x x x x M1 A1 Correct form on LHS and simplifies RHS. ( ) 2 2 1 1 + − = −+ y x x x C M1 A1 Integrates RHS. For M1, needs to integrate non-zero multiple of 2 . 1 − x x 3 4 2 C = + leading to 5 4 C = M1 Substitutes initial conditions. 5 2 4 2 1 1 −+ = + − x y x x M1 A1 Divides through by coefficient of y. 7

This question in 9231/23 Oct/Nov 2021

3 The variables t and x are related by the differential equation d 2 x d x 2 + + x = t + 1. 2 d t d t (a) Find the general solution for x in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … d 2 x(b) Deduce an approximate value of 2 for large positive values of t. [2] d t … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) 2 3 2 1 2 1 0 leadin i g to z z m     M1 Auxiliary equation.   1 2 3 3 2 2 e sin cos t x A t B t    A1 Complimentary function. 2 leading to leading to 2 2         x pt q x p x pt qt r B1 Particular integral and its derivatives. 2 2 2 2 1 p pt q pt qt r t        M1 Substitutes and equates coefficients. PI must have the correct form. 1, 2, 1 r q p    A1   1 2 3 2 3 2 2 e sin cos 2 1 t x A t B t t t       A1 6 3(b) 1 2 e 0 t   as t  M1 2 x   A1 FT 2

This question in 9231/21 May/June 2022

6 Use the substitution y = vx to find the solution of the differential equation d y 2 2 x = y + 9x + y d x for which y = 0 when x = 1. Give your answer in the form y = f ( x) , where f ( x) is a polynomial in x. [10] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6 d d d d y v v x x x   2 2 2 d 9 d v x v x vx x x v x           M1 Substitutes and derives first order separable equation. 2 d 9 d v x v x   A1 2 1 1 d d 9 v x x v     M1 Separates variables and integrates both sides. 1 sinh ln 3 v x C        A1 0 C  B1 Substitutes initial conditions.   2 3 9 ln 1 ln v v x    M1 Converts to logarithmic form. 2 2 2 9 3 y y x x    A1 4 2 2 2 2 9 6 9 0 x x y y y x      M1 Squares to eliminate radical. 2 2 3 ( 1) y x   A1 10

This question in 9231/21 May/June 2022

3 The variables t and x are related by the differential equation d 2 x d x 2 + + x = t + 1. 2 d t d t (a) Find the general solution for x in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … d 2 x(b) Deduce an approximate value of 2 for large positive values of t. [2] d t … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) 2 3 2 1 2 1 0 leadin i g to z z m     M1 Auxiliary equation.   1 2 3 3 2 2 e sin cos t x A t B t    A1 Complimentary function. 2 leading to leading to 2 2         x pt q x p x pt qt r B1 Particular integral and its derivatives. 2 2 2 2 1 p pt q pt qt r t        M1 Substitutes and equates coefficients. PI must have the correct form. 1, 2, 1 r q p    A1   1 2 3 2 3 2 2 e sin cos 2 1 t x A t B t t t       A1 6 3(b) 1 2 e 0 t   as t  M1 2 x   A1 FT 2

This question in 9231/22 May/June 2022

6 Use the substitution y = vx to find the solution of the differential equation d y 2 2 x = y + 9x + y d x for which y = 0 when x = 1. Give your answer in the form y = f ( x) , where f ( x) is a polynomial in x. [10] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6 d d d d y v v x x x   2 2 2 d 9 d v x v x vx x x v x           M1 Substitutes and derives first order separable equation. 2 d 9 d v x v x   A1 2 1 1 d d 9 v x x v     M1 Separates variables and integrates both sides. 1 sinh ln 3 v x C        A1 0 C  B1 Substitutes initial conditions.   2 3 9 ln 1 ln v v x    M1 Converts to logarithmic form. 2 2 2 9 3 y y x x    A1 4 2 2 2 2 9 6 9 0 x x y y y x      M1 Squares to eliminate radical. 2 2 3 ( 1) y x   A1 10

This question in 9231/22 May/June 2022

5 Find the particular solution of the differential equation d 2 y dy 2 2 2 + 2 + y = 4x + 3x + 3 , d x dx dy given that, when x = 0 , y = = 0 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 2m2 + 2m + 1 = 0  m = − 12  12 i M1 Correct auxiliary equation. − 12 x 1 1 A1 Complementary function. Accept “ y = ” y = e ( A cos 2 x + B sin 2 x ) missing. y = px 2 + qx + r leading to y ' = 2 px + q leading to y '' = 2 p B1 Particular integral and its derivatives. p = 4 4 p + q = 3 4 p + 2q + r = 3 M1 Substitutes and equates coefficients. q = −13 r = 13 A1 − 12 x 1 1 2 A1 General solution. y = e ( A cos 2 x + B sin 2 x ) + 4 x − 13 x + 13 − 12 x 1 1 1 1 1 − 12 x 1 1 M1 Full use of product rule. y must be of form y ' = e ( − 2 A sin 2 x + 2 B cos 2 x ) − 2 e ( A cos 2 x + B sin 2 x ) + 8 x − 13 shown in the general solution above with p  0. A+ 13 = 0 12 B − 12 A − 13 = 0 A = −13, B = 13 M1 A1 Uses initial conditions to derive two linear equations in two unknowns. − 12 x 1 1 2 A1 y = e ( −13cos 2 x + 13sin 2 x ) + 4 x − 13 x + 13 10

This question in 9231/21 Oct/Nov 2022

8 (a) Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di . [3] y 2 1 - ( i - 1) … … … … … … … (b) Find the solution of the differential equation d y 2 -1 i - y = i sin ( i - 1 ) , di where 0 1 i 1 2 , given that y = 1 when i = 1. Give your answer in the form y = f ( i) . [11] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) du 1 B1 −2(− 1) leading to (− 1) d = − 2 du d=  − 1 1 1 2 M1 A1 Applies substitution. For M1, integrand must be du = − u + C = − 1 − (− 1) + C d= − 2   of the form k , where k  0 is constant. For u  u  1 − (− 1) 2 A1, allow “ +C ” missing. 3 8(b) dy y −1 B1 Divides through by . − = sin (− 1) d  − −1 d − ln −1 M1 A1 Finds integrating factor. e = e =  d −1 −1 M1 Correct form on LHS using their integrating  y = sin (− 1) ( ) factor. d −1 −1   M1 A1 Integrates s in −1 (− 1) .  y = sin (− 1) − d 2  1 − (− 1)    − 1  1 M1 Applies part (a) and uses  2 d=  2 d+  2 d 1 dx = sin −1 x + C . 2  1 − (− 1)  1 − (− 1)  1 − (− 1)  1 − x −1 −1 2 −1 A1  y = sin (− 1) + 1 − (− 1) − sin (− 1) + C 8(b) 1 = 1 + C M1 Substitutes initial conditions into their expression in y and . −1 2 M1 A1 Divides through by their integrating factor. y = ( − 1) sin (− 1) +  1 − (− 1) 11

This question in 9231/21 Oct/Nov 2022

4 Find the solution of the differential equation 2 dx 2 ( 4t - 1 ) + 4x = 4t - 1 dt for which x = 3 when t = 1. Give your answer in the form x = f( t) . [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4 dx 4 B1 Divides through by 4t 2 − 1 . + x = 1 dt 4t 2 − 1 4 2 2 4 4 1 B1 Writes as partial fractions or a form to which the = − or = = 1 4t 2 − 1 2t − 1 2t + 1 4t 2 − 1 ( 2t ) 2 − 1 t 2 − ( 2 ) 2 formula from MF19 can be applied.  4 ( 4 t 2 −1) −1 dt ln ( 2 t −−1) ln(2 t +1) 2t − 1 M1 A1 Finds integrating factor. e = e = 2t + 1 d  2t − 1 2t − 1 2t + 1 − 2 M1 Correct form on LHS and attempt to integrate RHS.  x = = at + b dt  2t + 1  2t + 1 2t + 1 RHS must be of the form with a, b, c, d  0. ct + d 2t 1 A1 x −= t − ln(2t + 1) + C 2t + 1 1 = 1 − ln3 +C M1 Substitutes into their expression in x and t.  2t + 1    3   M1 A1 After integrating, divides their expression through x =    t + ln    by their coefficient of x. (Their coefficient must  2t − 1    2t + 1   depend on t.) 9

This question in 9231/22 Oct/Nov 2022

8 It is given that y = cosh u , where u 2 0 , and 2 2 2 d u du d u -x cosh u - 1 2 + + cosh u - 2 cosh u = 4e . e dx dx o e d x o (a) Show that d 2 y dy -x + - 2y = 4 e . [4] 2 dx dx … … … … … … … … … du (b) Find u in terms of x, given that, when x = 0 , u = ln 3 and = 3 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) dy du B1 = sinh u dx dx 2 2 2 B1 d y d u  du  = sinh u + cosh u 2 2   dx dx  dx  2 2 2 M1 Uses substitution to find y-x equation, AG. d y dy d u  du  du + − 2 y = sinh u + cosh u + sinh u − 2cosh u 2 2   dx dx dx  dx  dx 2  d 2 u du   du  2 − x A1 AG cosh u − 1  2 +  + cosh u   − 2cosh u = 4e  dx dx   dx  8(a) Alternative for 8(a) du −1 dy B1 = ( sinh u ) dx dx d 2 u  1  d 2 y d y  cosh u  d u B1 2 =   2 −  2  d x  sinh u  d x d x  sinh u  d x d 2 u  1  d 2 y  dy 2 cosh u  = − 2   2   3  dx  sinh u  dx  dx  sinh u    1  d 2 y  dy 2 cosh u  −1 dy  cosh u  dy  2 M1 Uses substitution to find y-x equation, AG. 2cosh u sinh u    2 −   3  + ( sinh u )  + 2   −   sinh u  dx  dx  sinh u  dx  sinh u  dx    d 2 y  dy 2 cosh u  dy cosh u  dy  2 − + + − 2cosh u 2   2  2   dx  dx  sinh u  dx sinh u  dx  d 2 y dy − x A1 AG + − 2 y = 4e dx 2 d x 4 8(b) m 2 + m − 2 = 0  m = 1, − 2 M1 Auxiliary equation. y = A e x + B e− 2 x A1 Complimentary function. Allow ‘ y = ’ missing. y = ke − x y ' = − ke − x y '' = ke − x B1 Particular integral and its derivatives. ke − x − ke − x − 2 ke − x = 4e − x M1 Substitutes and equates coefficients. k = −2 A1 WWW y = cosh u = Ae x + Be −2 x − 2e − x A1 Must have ‘ y = ’ or ‘cosh u = ’ . du x −2 x − x B1 y ' = sinh u = Ae − 2 Be + 2e dx A + B − 2 = 53 A − 2B + 2 = 4 A = 289 , B = 95 M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, allow substitution into their y and 'y if two linear equations in two unknowns derived. −1 28 x 5 −2 x − x A1 Substitutes for y and finds u in terms of x. u = cosh ( 9 e + 9 e − 2e ) 10

This question in 9231/22 Oct/Nov 2022

5 Find the particular solution of the differential equation d 2 y dy 2 2 2 + 2 + y = 4x + 3x + 3 , d x dx dy given that, when x = 0 , y = = 0 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 2m2 + 2m + 1 = 0  m = − 12  12 i M1 Correct auxiliary equation. − 12 x 1 1 A1 Complementary function. Accept “ y = ” y = e ( A cos 2 x + B sin 2 x ) missing. y = px 2 + qx + r leading to y ' = 2 px + q leading to y '' = 2 p B1 Particular integral and its derivatives. p = 4 4 p + q = 3 4 p + 2q + r = 3 M1 Substitutes and equates coefficients. q = −13 r = 13 A1 − 12 x 1 1 2 A1 General solution. y = e ( A cos 2 x + B sin 2 x ) + 4 x − 13 x + 13 − 12 x 1 1 1 1 1 − 12 x 1 1 M1 Full use of product rule. y must be of form y ' = e ( − 2 A sin 2 x + 2 B cos 2 x ) − 2 e ( A cos 2 x + B sin 2 x ) + 8 x − 13 shown in the general solution above with p  0. A+ 13 = 0 12 B − 12 A − 13 = 0 A = −13, B = 13 M1 A1 Uses initial conditions to derive two linear equations in two unknowns. − 12 x 1 1 2 A1 y = e ( −13cos 2 x + 13sin 2 x ) + 4 x − 13 x + 13 10

This question in 9231/23 Oct/Nov 2022

8 (a) Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di . [3] y 2 1 - ( i - 1) … … … … … … … (b) Find the solution of the differential equation d y 2 -1 i - y = i sin ( i - 1 ) , di where 0 1 i 1 2 , given that y = 1 when i = 1. Give your answer in the form y = f ( i) . [11] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) du 1 B1 −2(− 1) leading to (− 1) d = − 2 du d=  − 1 1 1 2 M1 A1 Applies substitution. For M1, integrand must be du = − u + C = − 1 − (− 1) + C d= − 2   of the form k , where k  0 is constant. For u  u  1 − (− 1) 2 A1, allow “ +C ” missing. 3 8(b) dy y −1 B1 Divides through by . − = sin (− 1) d  − −1 d − ln −1 M1 A1 Finds integrating factor. e = e =  d −1 −1 M1 Correct form on LHS using their integrating  y = sin (− 1) ( ) factor. d −1 −1   M1 A1 Integrates s in −1 (− 1) .  y = sin (− 1) − d 2  1 − (− 1)    − 1  1 M1 Applies part (a) and uses  2 d=  2 d+  2 d 1 dx = sin −1 x + C . 2  1 − (− 1)  1 − (− 1)  1 − (− 1)  1 − x −1 −1 2 −1 A1  y = sin (− 1) + 1 − (− 1) − sin (− 1) + C 8(b) 1 = 1 + C M1 Substitutes initial conditions into their expression in y and . −1 2 M1 A1 Divides through by their integrating factor. y = ( − 1) sin (− 1) +  1 − (− 1) 11

This question in 9231/23 Oct/Nov 2022

2 Use the substitution z = x + y to find the solution of the differential equation d y 1 + 3x + 3y = d x 3x + 3y - 1 for which y = 0 when x = 1. Give your answer in the form a ln ( x + y) + b ( x - y) + c = 0 , where a, b and c are constants to be determined. [7] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 2 d dz 1 d d y x x   B1 d 1 3 1 d 3 1 z z x z    M1 Substitutes and derives first order separable equation. d 1 3 6 1 d 3 1 3 1 z z z x z z      A1 1 1 1 2 6 d 1d z z x      M1 Separates variables and integrates both sides. 1 1 1 1 1 2 6 2 2 6 ln ln( ) z z x C x y x y C         A1 1 2 C  M1 Substitutes initial conditions into their expression.   1 1 1 6 2 2 ln( ) 0 x y x y      A1 OE. 7

This question in 9231/21 May/June 2023

6 Find the particular solution of the differential equation d 2 x d x - 12 + 36 x = 37 sin t , 2 d t d t dx given that, when t = 0, x = = 0 . [11] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6 2 12 36 0 m m    M1 Auxiliary equation.   6e t x At B   A1 Complementary function. Allow “ x ” missing. sin cos ' cos sin '' sin cos x p t q t x p t q t x p t q t         M1 A1 Particular integral and its derivatives. 12 36 37 p q p     12 36 0 q p q    M1 Substitutes and equates coefficients. Must have two unknowns. 35 37 p  12 37 q  A1   35 12 37 6 37 e sin cos t x At B t t     A1 FT Must have “ x ”. FT on CF.   35 12 37 37 6 6 ' e 6e cos sin t t x A At B t t      M1* Differentiates using product rule. Must have their PI differentiated. 12 37 B  35 37 6 0 1 A B A      DM1 A1 Forms simultaneous equations using initial conditions.   35 12 12 37 37 37 6e sin cos t x t t t     A1 Must have “ x ”. 11

This question in 9231/21 May/June 2023

2 Use the substitution z = x + y to find the solution of the differential equation d y 1 + 3x + 3y = d x 3x + 3y - 1 for which y = 0 when x = 1. Give your answer in the form a ln ( x + y) + b ( x - y) + c = 0 , where a, b and c are constants to be determined. [7] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 2 d dz 1 d d y x x   B1 d 1 3 1 d 3 1 z z x z    M1 Substitutes and derives first order separable equation. d 1 3 6 1 d 3 1 3 1 z z z x z z      A1 1 1 1 2 6 d 1d z z x      M1 Separates variables and integrates both sides. 1 1 1 1 1 2 6 2 2 6 ln ln( ) z z x C x y x y C         A1 1 2 C  M1 Substitutes initial conditions into their expression.   1 1 1 6 2 2 ln( ) 0 x y x y      A1 OE. 7

This question in 9231/22 May/June 2023

6 Find the particular solution of the differential equation d 2 x d x - 12 + 36 x = 37 sin t , 2 d t d t dx given that, when t = 0, x = = 0 . [11] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6 2 12 36 0 m m    M1 Auxiliary equation.   6e t x At B   A1 Complementary function. Allow “ x ” missing. sin cos ' cos sin '' sin cos x p t q t x p t q t x p t q t         M1 A1 Particular integral and its derivatives. 12 36 37 p q p     12 36 0 q p q    M1 Substitutes and equates coefficients. Must have two unknowns. 35 37 p  12 37 q  A1   35 12 37 6 37 e sin cos t x At B t t     A1 FT Must have “ x ”. FT on CF.   35 12 37 37 6 6 ' e 6e cos sin t t x A At B t t      M1* Differentiates using product rule. Must have their PI differentiated. 12 37 B  35 37 6 0 1 A B A      DM1 A1 Forms simultaneous equations using initial conditions.   35 12 12 37 37 37 6e sin cos t x t t t     A1 Must have “ x ”. 11

This question in 9231/22 May/June 2023

2 The variables x and y are related by the differential equation d 2 x dx 2 6 2 + 5 + x = t + 10 t + 13 . dt dt (a) Find the general solution for x in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … (b) State an approximate solution for large positive values of t. [1] … …

7 marks

Mark scheme: 2(a) 2 3 1 6 , 5 1 1 0 2 m m m      M1 Auxiliary equation. 1 1 3 2 e e t t x A B     A1 Complementary function. Allow with ‘x =’ missing. 2 ' 2 '' 2 x p qt rt x q rt x r         B1 Particular integral and its derivatives. 2 2 12 5 10 10 13 r q rt p qt rt t t         M1 Substitutes and equates coefficients. 1, 0, 1 r q p    A1 1 1 3 2 2 1 e e t t x A B t       A1 Must have x . 6 2(b) 2 1 x t   B1FT Must have ‘x =’. Accept ‘x is approximately’ but do not accept ‘ x ‘. FT on their PI. 1

This question in 9231/23 May/June 2023

5 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that 2 cosh 2 x = cosh 2 x + 1. [3] … … … … … … … … … … … … (b) Find the solution of the differential equation dy + 2y tanh x = 1 dx for which y = 1 when x = 0 . Give your answer in the form y = f ( x) . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5(a) 1 2 cosh e e x x x    B1     2 2 1 2 2 2 4 e e e e 2 cosh 2 1 x x x x x         M1 A1 Expands, AG. 3 Question Answer Partial Marks Guidance 5(b) d 2lncosh 2 tanh 2 e e cosh x x x x    M1 A1 Finds integrating factor.   2 2 d cosh cosh d y x x x  M1 Correct form on LHS,   d d yI x for their integrating factor I, and attempt to integrate their RHS.   2 2 1 1 1 cosh cosh d cosh2 1d sinh2 2 2 2 y x x x x x x x C         M1 A1 Uses   2 1 cosh cosh 2 1 . 2 x x   M1 A1 is for RHS. 1 C  M1 Finds C.   2 1 1 sech sinh2 1 4 2 y x x x    M1 A1 Divides through by their coefficient of y. 8

This question in 9231/23 May/June 2023

4 Find the particular solution of the differential equation d 2 y dy 2 + 2 + 3 y = 27 x , 2 d x dx dy given that, when x = 0 , y = 2 and = - 8 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 4 m 2 + 2 m + 3 = 0 M1 Auxiliary equation. A1 Complementary function. Allow with “ y = ” missing. [ y =]e − x A cos 2 x + B sin 2 x ( ) y = px 2 + qx + r  y ' = 2 px + q  y '' = 2 p B1 Particular integral and its derivatives. 3 p = 27 3q + 4 p = 0 2 p + 2q + 3r = 0 M1 Substitutes and equates coefficients. 4 p = 9 q = −12 r = 2 A1 y = e − x A cos 2 x + B sin 2 x + 9 x 2 − 12 x + 2 A1 General solution. Must have “ y = ”. ( ) − x − x M1* Differentiates. Must use product rule. y ' = e − 2 A sin 2 x + 2 B cos 2 x − e A cos 2 x + B sin 2 x + 18 x − 12 ( ) ( ) A + 2 = 2 2 B − A − 12 = −8 DM1 Uses initial conditions A1  A = 0, B = 2 2 y = 2 2e − x sin 2 x + 9 x 2 − 12 x + 2 A1 Must have “ y = ”. 10

This question in 9231/21 Oct/Nov 2023

6 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . [3] … … … … … … … … … (b) Using the substitution u = sinh x , find sinh 2 2x cosh x dx . [4] y … … … … … … … … … … … … … … … … (c) Find the particular solution of the differential equation d y 2 + y tanh x = sinh 2x , d x given that y = 4 when x = 0 . Give your answer in the form y = f ( x) . [7] … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 6(a) 1 x − x 1 x − x B1 cosh x = e + e sinh x = e − e ( ) ( ) 2 2 1 x − x x − x 1 2 x −2 x M1 A1 Expands, AG. e − e e +e = e − e = sinh2 x ( )( ) ( ) 2 2 3 6(b) u = sinh x  du = cosh x dx B1 2 2 2 2 2 M1 Applies identities to find integral in terms of u.  sinh 2 x cosh x dx = 4  sinh x cosh x du = 4  sinh x ( sinh x + 1) du = 4  u 2 u 2 + 1 du A1 ( )  1 5 1 3   1 5 1 3  A1 = 4 u + u ( + C ) = 4 sinh x + sinh x ( + C )      5 3   5 3  4 tanh xdx lncosh x M1 A1 Finds integrating factor.6(c)  e = e = cosh x d 2 M1 Correct form on LHS and attempt to integrate RHS. ( y cosh x ) = sinh 2 x cosh x dx  1 5 1 3  M1 A1 Integrates RHS using their part (b). y cosh x = 4 sinh x + sinh x + C    5 3  4 = C M1 Substitutes initial conditions.  1 5 1 3  A1 y = 4sech x sinh x + sinh x + 1    5 3  7

This question in 9231/21 Oct/Nov 2023

4 Find the solution of the differential equation dy + 3 y = sin x dx for which y = 1 when x = 0 . Give your answer in the form y = f ( x) . [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4  3 dx 3 x M1 A1 Finds integrating factor. e = e d 3 x 3 x M1 Correct form on LHS and attempt ye = e sin x ( ) to integrate RHS. dx EITHER M1 A1 Integrates by parts once or uses  e3 x sin x dx = −e 3 x cos x + 3 e 3 x cos x dx sin x = e i x − e -i x . 2i OR e 3 x cos x dx  e3 x sin x dx = 13 e 3 x sin x − 13  EITHER M1 Integrates by parts again or  e 3 x sin x dx = − e 3 x cos x + 3 ( e 3 x sin x − 3  e 3 x sin x dx ) substitutes e ix − e -ix = sin x and 2i OR ix -ix e + e 3 x 1 3 x 1 1 3 x 1 3 x = cos x. e sin x dx e sin x dx = 3 e sin x − 3 ( 3  e cos x + 3  ) 2 ye3 x = 110 e3 x ( 3sin x − cos x ) + C A1 Must not see i. 1 = − 101 + C M1 Finds C. Substitutes into their expression (must be integrated). y = 103 sin x − 101 cos x + 1011 e−3 x A1 Divides through by coefficient of y. 9

This question in 9231/22 Oct/Nov 2023

8 It is given that v = y4 and 3 d 2 y 2 dy 2 3 dy 4 -2 x y 2 + 3 y + y + y = e . dx e dx o dx (a) Show that d 2 v dv -2 x + + 4v = 4 e . [4] 2 dx dx … … … … … … … … … … … … … … … … … … … … … … … dy 3(b) Find y in terms of x, given that, when x = 0 , y = 1 and = - . [10] dx 8 … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) dv 3 dy B1 v = y 4 = 4 y dx dx 2 2 2 B1 d v 3 d y 2  dy  = 4 y + 12 y 2 2   dx dx  dx  2 2 2 M1 Uses substitution to find v-x d v dv 3 d y 2  dy  3 dy 4 + + 4v = 4 y + 12 y + 4 y + 4 y equation, AG.   2 2 dx dx dx  dx  dx  3 d 2 y 2  dy  2 3 dy 4  −2 x A1 AG. = 4  y 2 + 3 y   + y + y  = 4e  dx  dx  dx    Alternative method for question 8(a) dy 1 dv 1 dv (B1) y = v 1 = 4 = 3 3 4 dx dx 4 y dx 4v 2 2 2 (B1) d y 1 d v 3  dv  2 = 3 2 − 7   4  dx  dx 4v 4 dx 16v (M1) Uses substitution to find v-x 3 1  1 d 2 v 3  dv  2  3  1 dv  −2 x 4 2  1 dv  2 4 v  3 2 − 7    + 3v  3  + v  3  + v = e equation, AG. 4    dx  4 dx   4v 4 dx   4v 4 dx 16v   4v 2 2 2 (A1) 1 d v 3  dv  3  dv  1 dv −2 x − + + + v = e 2     4 dx 16v  dx  16v  dx  4 dx 4 8(b) m 2 + m + 4 = 0 M1 Auxiliary equation.  v =  e − 12 x ( 15 15 A1 Complimentary function. Allow A cos 2 x + B sin 2 x ) ‘v =’ missing. v = ke −2 x  v ' = −2ke −2 x  v '' = 4ke −2 x B1 Particular integral and its derivatives. 4 ke −2 x − 2 ke −2 x + 4 ke −2 x = 4e −2 x M1 Substitutes and equates coefficients. k = 23 A1 4 − 12 x 15 15 2 −2 x A1 Substitutes for y and find v in terms v = y = e + 3 e ( A cos 2 x + B sin 2 x ) of x. Must not see i. 1 B1 3 dy − 2 x 15 15 15 15 1 − 12 x 15 15 4 −2 x v ' = 4 y = e − 3 e ( − 2 A sin 2 x + 2 B cos 2 x ) − 2 e ( A cos 2 x + B sin 2 x ) dx 2 3 15 1 4 1 M1 A1 Substitutes initial conditions and 1 = A + 3 −=2 2 B − 2 A − 3  A = 3 , B = 0 forms simultaneous equations. Must have used the product rule when differentiating their v for M1. 1 A1 Accept 1 − 12 x 15 2 −2 x 4 1 y = 1 ( 3 e cos 2 x + 3 e ) 1 − 2 x 15 2 −2 x y =  4 ( 3 e cos 2 x + 3 e ) 10

This question in 9231/22 Oct/Nov 2023

4 Find the particular solution of the differential equation d 2 y dy 2 + 2 + 3 y = 27 x , 2 d x dx dy given that, when x = 0 , y = 2 and = - 8 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 4 m 2 + 2 m + 3 = 0 M1 Auxiliary equation. A1 Complementary function. Allow with “ y = ” missing. [ y =]e − x A cos 2 x + B sin 2 x ( ) y = px 2 + qx + r  y ' = 2 px + q  y '' = 2 p B1 Particular integral and its derivatives. 3 p = 27 3q + 4 p = 0 2 p + 2q + 3r = 0 M1 Substitutes and equates coefficients. 4 p = 9 q = −12 r = 2 A1 y = e − x A cos 2 x + B sin 2 x + 9 x 2 − 12 x + 2 A1 General solution. Must have “ y = ”. ( ) − x − x M1* Differentiates. Must use product rule. y ' = e − 2 A sin 2 x + 2 B cos 2 x − e A cos 2 x + B sin 2 x + 18 x − 12 ( ) ( ) A + 2 = 2 2 B − A − 12 = −8 DM1 Uses initial conditions A1  A = 0, B = 2 2 y = 2 2e − x sin 2 x + 9 x 2 − 12 x + 2 A1 Must have “ y = ”. 10

This question in 9231/23 Oct/Nov 2023

6 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . [3] … … … … … … … … … (b) Using the substitution u = sinh x , find sinh 2 2x cosh x dx . [4] y … … … … … … … … … … … … … … … … (c) Find the particular solution of the differential equation d y 2 + y tanh x = sinh 2x , d x given that y = 4 when x = 0 . Give your answer in the form y = f ( x) . [7] … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 6(a) 1 x − x 1 x − x B1 cosh x = e + e sinh x = e − e ( ) ( ) 2 2 1 x − x x − x 1 2 x −2 x M1 A1 Expands, AG. e − e e +e = e − e = sinh2 x ( )( ) ( ) 2 2 3 6(b) u = sinh x  du = cosh x dx B1 2 2 2 2 2 M1 Applies identities to find integral in terms of u.  sinh 2 x cosh x dx = 4  sinh x cosh x du = 4  sinh x ( sinh x + 1) du = 4  u 2 u 2 + 1 du A1 ( )  1 5 1 3   1 5 1 3  A1 = 4 u + u ( + C ) = 4 sinh x + sinh x ( + C )      5 3   5 3  4 tanh xdx lncosh x M1 A1 Finds integrating factor.6(c)  e = e = cosh x d 2 M1 Correct form on LHS and attempt to integrate RHS. ( y cosh x ) = sinh 2 x cosh x dx  1 5 1 3  M1 A1 Integrates RHS using their part (b). y cosh x = 4 sinh x + sinh x + C    5 3  4 = C M1 Substitutes initial conditions.  1 5 1 3  A1 y = 4sech x sinh x + sinh x + 1    5 3  7

This question in 9231/23 Oct/Nov 2023

1 6 (a) Show that cosh x + sinh x 2 = e 2 x . [2] … … … … … … … … … … … … … (b) Find the particular solution of the differential equation 1 d 2 y dy 2 + + 3 y = 5 cosh x + sinh x , 2 dx dx ` j dy 4 given that, when x = 0 , y = 1 and = . [10] dx 3 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a)       1 1 2 2 1 1 2 2 cosh sinh e e e e x x x x x x         M1 Substitutes sinh and cosh in terms of exponentials.    1 2 1 2 e e x x  A1 AG 2 Question Answer Marks Guidance 6(b) 1 2 2 1 2 3 i 0 11 m m m        M1 Auxiliary equation.   1 2 11 11 2 2 e cos sin x y A x B x    A1 Complimentary function. Allow ‘ y ’ missing. 1 1 1 2 2 2 1 1 2 4 e ' e '' e x x x y k y k y k      B1 Particular integral and its derivatives. 1 1 1 2 2 1 2 2 1 1 1 1 4 2 4 2 e e 3 e 5e 3 5 x x x x k k k k k k        M1 Substitutes and equates coefficients. PI must be correct form ( bx ae ). 4 3 k  A1   2 2 1 1 11 11 4 2 2 3 e cos sin e x x y A x B x     A1 FT Must have ‘ y ’. FT on their CF.     2 1 1 2 2 1 11 11 11 11 11 11 1 2 2 2 2 2 2 2 2 3 d e sin cos e cos sin e d x x x y A x B x A x B x x         B1 4 3 1 A   11 4 1 2 3 2 2 3 B A    1 1 3 11 , A B    M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form   2 2 1 1 11 11 1 1 4 3 2 2 3 11 e cos sin e x x y x x      A1 Must have ‘ y ’. 10

This question in 9231/21 May/June 2024

7 (a) Use the substitution u = 1 + x 2 to find x dx . [2] y 2 1 + x … … … … … … … … … … … … … (b) Find the solution of the differential equation dy 2 -1 x - y = x sinh x , dx given that y = 1 when x = 1. Give your answer in the form y = f x [10] ` j. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 7(a)     1 2 2 2 1 d d 1 1 x x u u C x C u x             2 7(b) 1 s d d inh y y x x x x    B1 Divides through by .x 1 1 ln d e e x x x x        M1 A1 Finds integrating factor.   1 1 d sinh d x y x x    M1 Correct form on LHS.   d d Iy x for their integrating factor I. 1 1 2 sinh d 1 x y x x x x x       M1 A1 Integrates RHS. RHS must be of the form 1 sinh c x  . 1 1 2 sinh 1 x y x x C x       A1   1 2 sinh ln 1 x x x       1 ln 1 2 2 C     *M1 Substitutes initial conditions.     2 1 2 1 sinh ln 1 2 1 2 y x x x x x x         DM1 A1 Divides through by their integrating factor. Accept   2 1 2 1 1 sinh 1 1 2 . sinh y x x x x x x         10

This question in 9231/21 May/June 2024

1 6 (a) Show that cosh x + sinh x 2 = e 2 x . [2] … … … … … … … … … … … … … (b) Find the particular solution of the differential equation 1 d 2 y dy 2 + + 3 y = 5 cosh x + sinh x , 2 dx dx ` j dy 4 given that, when x = 0 , y = 1 and = . [10] dx 3 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a)       1 1 2 2 1 1 2 2 cosh sinh e e e e x x x x x x         M1 Substitutes sinh and cosh in terms of exponentials.    1 2 1 2 e e x x  A1 AG 2 Question Answer Marks Guidance 6(b) 1 2 2 1 2 3 i 0 11 m m m        M1 Auxiliary equation.   1 2 11 11 2 2 e cos sin x y A x B x    A1 Complimentary function. Allow ‘ y ’ missing. 1 1 1 2 2 2 1 1 2 4 e ' e '' e x x x y k y k y k      B1 Particular integral and its derivatives. 1 1 1 2 2 1 2 2 1 1 1 1 4 2 4 2 e e 3 e 5e 3 5 x x x x k k k k k k        M1 Substitutes and equates coefficients. PI must be correct form ( bx ae ). 4 3 k  A1   2 2 1 1 11 11 4 2 2 3 e cos sin e x x y A x B x     A1 FT Must have ‘ y ’. FT on their CF.     2 1 1 2 2 1 11 11 11 11 11 11 1 2 2 2 2 2 2 2 2 3 d e sin cos e cos sin e d x x x y A x B x A x B x x         B1 4 3 1 A   11 4 1 2 3 2 2 3 B A    1 1 3 11 , A B    M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form   2 2 1 1 11 11 1 1 4 3 2 2 3 11 e cos sin e x x y x x      A1 Must have ‘ y ’. 10

This question in 9231/22 May/June 2024

7 (a) Use the substitution u = 1 + x 2 to find x dx . [2] y 2 1 + x … … … … … … … … … … … … … (b) Find the solution of the differential equation dy 2 -1 x - y = x sinh x , dx given that y = 1 when x = 1. Give your answer in the form y = f x [10] ` j. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 7(a)     1 2 2 2 1 d d 1 1 x x u u C x C u x             2 7(b) 1 s d d inh y y x x x x    B1 Divides through by .x 1 1 ln d e e x x x x        M1 A1 Finds integrating factor.   1 1 d sinh d x y x x    M1 Correct form on LHS.   d d Iy x for their integrating factor I. 1 1 2 sinh d 1 x y x x x x x       M1 A1 Integrates RHS. RHS must be of the form 1 sinh c x  . 1 1 2 sinh 1 x y x x C x       A1   1 2 sinh ln 1 x x x       1 ln 1 2 2 C     *M1 Substitutes initial conditions.     2 1 2 1 sinh ln 1 2 1 2 y x x x x x x         DM1 A1 Divides through by their integrating factor. Accept   2 1 2 1 1 sinh 1 1 2 . sinh y x x x x x x         10

This question in 9231/22 May/June 2024

5 Find the particular solution of the differential equation d 2 x d x 2 6 - 5 + x = t + t + 1, d t 2 d t dx given that, when t = 0 , x = 12 and = -6 . [10] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 6m2 − 5m + 1 = 0  m = 12 , 13 M1 Auxiliary equation, two distinct real roots. 1 t 13 t A1 Complementary function. x = A2e + Be x = pt 2 + qt + r  x ' = 2 pt + q  x '' = 2 p B1 Particular integral and its derivatives. p = 1 −10 p + q = 1 12 p − 5q + r = 1 M1 Substitutes and equates coefficients. p = 1 q = 11 r = 44 A1 1 2 t 13 t 2 A1 General solution. Need to see ‘ x = ’. FT on CF. x = Ae + Be + t + 11t + 44 1 12 t 1 13 t M1 Differentiates. x ' = 2 Ae + 3 Be + 2t + 11 A + B + 44 = 12 12 A + 13 B + 11 = − 6  A = −38, B = 6 M1A1 Uses initial conditions. x = −38e 1 2 t + 6e 1 3 t + t 2 + 11t + 44 A1 Need to see ‘ x = ’. 10

This question in 9231/21 Oct/Nov 2024

7 (a) Show that an appropriate integrating factor for 2 dy 2 x + 16 + y = x x + 16 dx is 1 x + 1 x 2 + 16 . [4] 4 4 … … … … … … … … … … … … … (b) Hence find the solution of the differential equation 2 dy 2 x + 16 + y = x x + 16 dx for which y = 6 when x = 3 . [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) d y y B1 2 + = x Divides through by x + 16 . d x x 2 + 16 1 M1A1 Finds integrating factor dx x  ) sinh −1 ( 4 x 2 +16 e = e 1 1 2 A1 AG. = 4 x + 4 x + 16 4 7(b) d 2 2 2 M1A1 Correct form on LHS and RHS. y x + x + 16 = x + x x + 16 dx ( ( ) ) 3 2 1 3 1 2 2 M1A1 Integrates RHS. RHS of the correct form. y x + x + 16 x + 16 + C = 3 x + 3 ( ) ) ( 27 25 M1 Substitutes initial conditions into their expression. 6 3 + 25 = 3 + 3 25 + C ( ) 3 2 1 3 1 2 2 8 A1 OE. y x + x + 16 x + 16 − 3 = 3 x + 3 ( ) ) ( 6

This question in 9231/21 Oct/Nov 2024

5 Find the particular solution of the differential equation d 2 y dy 2 3 + 2 + y = x , dx 2 dx dy given that, when x = 0 , y = = 0 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 2 1 2 M1 Auxiliary equation. M1 for correct type of 3m + 2m + 1 = 0  m = −3 3 i roots (complex). − 13 x 2 2 A1 Complementary function. y = e A cos x + B sin x ( 3 3 ) y = px 2 + qx + r  y ' = 2 px + q  y '' = 2 p B1 Particular integral and its derivatives. p = 1 4 p + q = 0 6 p + 2q + r = 0 M1 Substitutes and equates coefficients. q = −4 r = 2 A1 − 13 x 2 2 2 A1FT General solution. Must have ‘ y = ’. FT on y = e A cos x + B sin x + x − 4 x + 2 ( 3 3 ) CF. − 13 x 2 2 2 2 1 − 13 x 2 2 *M1 Differentiates. GS must be in correct form. y ' = e − A sin x + B cos x − e A cos x + B sin x + 2 x − 4 ( 3 3 3 3 ) 3 ( 3 3 )  A = − 2, B = 5 2 A + 2 = 0 32 B − 13 A − 4 = 0 DM1A1 Uses initial conditions. − 13 x 2 2 2 A1 Must have ‘ y = ’. y = e −2cos x + 5 2sin x + x − 4 x + 2 ( 3 3 ) 10

This question in 9231/22 Oct/Nov 2024

7 (a) Show that ( ln ( tanh x)) = 2 cosech 2x . [3] dx … … … … … … … … … … … … … … … (b) Find the solution of the differential equation dy sinh 2x + 2 y = sinh 2 x dx for which y = 5 when x = ln 2 . Give your answer in an exact form. [7] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7(a) d sech 2 x M1A1 Applies chain rule. ( ln ( tanh x ) ) = dx tanh x 1 2 A1 AG. = = . sinh x cosh x sinh2 x 3 7(b) d y 2 y B1 Divides through by sinh(2 )x . + = 1 d x sinh(2 x ) d M1A1 Correct form on LHS and RHS. ( y tanh x ) = tanh x dx y tanh x = ln(cosh x ) + C M1A1 Integrates RHS. 3 5 ) = ln 5 ( 5 4 +C M1 Substitutes initial conditions. y tanh x = ln(cosh x ) + 3 − ln 54 A1 Accept equivalent exact form. 7

This question in 9231/22 Oct/Nov 2024

5 Find the particular solution of the differential equation d 2 x d x 2 6 - 5 + x = t + t + 1, d t 2 d t dx given that, when t = 0 , x = 12 and = -6 . [10] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 6m2 − 5m + 1 = 0  m = 12 , 13 M1 Auxiliary equation, two distinct real roots. 3 t A1 Complementary function. x = A2e1 t + Be 1 x = pt 2 + qt + r  x ' = 2 pt + q  x '' = 2 p B1 Particular integral and its derivatives. p = 1 −10 p + q = 1 12 p − 5q + r = 1 M1 Substitutes and equates coefficients. p = 1 q = 11 r = 44 A1 x = Ae 1 2 t + Be 1 3 t + t 2 + 11t + 44 A1 General solution. Need to see ‘ x = ’. FT on CF. 1 12 t 1 13 t M1 Differentiates. x ' = 2 Ae + 3 Be + 2t + 11 A + B + 44 = 12 12 A + 13 B + 11 = − 6  A = −38, B = 6 M1A1 Uses initial conditions. x = −38e 1 2 t + 6e 1 3 t + t 2 + 11t + 44 A1 Need to see ‘ x = ’. 10

This question in 9231/23 Oct/Nov 2024

7 (a) Show that an appropriate integrating factor for 2 dy 2 x + 16 + y = x x + 16 dx is 1 x + 1 x 2 + 16 . [4] 4 4 … … … … … … … … … … … … … (b) Hence find the solution of the differential equation 2 dy 2 x + 16 + y = x x + 16 dx for which y = 6 when x = 3 . [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) d y y B1 2 Divides through by x + 16 . + = x d x x 2 + 16 1 M1A1 Finds integrating factor dx x  ) sinh −1 ( 4 x 2 +16 e = e 1 1 2 A1 AG. = 4 x + 4 x + 16 4 7(b) d 2 2 2 M1A1 Correct form on LHS and RHS. y x + x + 16 = x + x x + 16 dx ( ( ) ) 3 2 1 3 1 2 2 M1A1 Integrates RHS. RHS of the correct form. y x + x + 16 x + 16 + C = 3 x + 3 ( ) ) ( 27 25 M1 Substitutes initial conditions into their expression. 6 3 + 25 = 3 + 3 25 + C ( ) 3 2 1 3 1 2 2 8 A1 OE. y x + x + 16 x + 16 − 3 = 3 x + 3 ( ) ) ( 6

This question in 9231/23 Oct/Nov 2024

5 Find the particular solution of the differential equation d 2 x dx -t 6 + 3 + 6x = e , dt 2 dt dx given that, when t = 0 , x = = 0 . [10] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 6 m 2 + 3m + 6 = 0  m = −14 14 i 15 M1 Auxiliary equation. M1 for correct type (complex) of roots. − 14 t 15 15 A1 Complimentary function. x = e ( A cos 4 t + B sin 4 t ) Allow ‘ x = ’ missing. Using x instead of t is A0. x = ke−t  x ' = −ke−t  x '' = ke −t B1 Particular integral and its derivatives. 6ke−t − 3ke−t + 6ke−t = e−t M1 Substitutes and equates coefficients. k = 19 A1 − 14 t 15 15 1 − t A1 FT General solution. FT on CF and their value of k. Must x = e + 9 e ( A cos 4 t + B sin 4 t ) have ‘ x = ’. 15 − 14 t 15 15 1 − 14 t 15 15 1 − t *M1 Using product rule. − 9 e x ' = 4 e ( − A sin 4 t + B cos 4 t ) − 4 e ( A cos 4 t + B sin 4 t ) 1 15 1 1 1 1 DM1 A1 Substitutes initial conditions and forms simultaneous 0 = A + 9 0 = 4 B − 4 A − 9  A = − 9 , B = 3 15 equations. 1 − 14 t 1 15 1 15 1 − t A1 Must have ‘ x = ’. + 9 e x = 3 e ( − 3 cos 4 t + 15 sin 4 t ) 10

This question in 9231/21 May/June 2025

7 Find the solution of the differential equation dy x + 5 - y = 1, dx x 2 + 10x + 61 given that y = 0 when x = 3 . Give your answer in an exact form. [10] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: Finds integrating factor using dx = ln f ( x ) . x + 10 x + 61 dx = 2 ln ( )  f ( x )7  x x + 5 1 2 M1 A1 f '( x ) 2 + 10 x + 61 − 12 ln ( x 2 +10 x + 61) 1 A1 e = x 2 + 10 x + 61 d   y = 1 M1 Correct form on LHS using their integrating factor.  2  2 dx  x + 10 x + 61  x + 10 x + 61 2 2 B1 Completes the square. x + 10 x + 61 = ( x + 5 ) + 36  1 −1  x + 5 *M1 A1 Integrates RHS. Attempting to integrate by parts is dx = sinh + C M0.    6  2 + 6 2  ( x + 5 ) 8 8 C = − sinh −1 ( 6 ) = − ln ( 6 + 53 ) = − ln3 DM1 A1 Substitutes initial conditions.  y −1  x + 5  −1 4 A1 y  x + 5 x 2 + 10 x + 61 = sinh   − sinh ( 3 ) Accept = ln  +  . 2  x + 10 x + 61  6  18 18  x 2 + 10 x + 61   10

This question in 9231/21 May/June 2025

5 Find the particular solution of the differential equation d 2 x dx -t 6 + 3 + 6x = e , dt 2 dt dx given that, when t = 0 , x = = 0 . [10] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5 6 m 2 + 3m + 6 = 0  m = −14 14 i 15 M1 Auxiliary equation. M1 for correct type (complex) of roots. − 14 t 15 15 A1 Complimentary function. x = e ( A cos 4 t + B sin 4 t ) Allow ‘ x = ’ missing. Using x instead of t is A0. x = ke−t  x ' = −ke−t  x '' = ke −t B1 Particular integral and its derivatives. 6ke−t − 3ke−t + 6ke−t = e−t M1 Substitutes and equates coefficients. k = 19 A1 − 14 t 15 15 1 − t A1 FT General solution. FT on CF and their value of k. Must x = e + 9 e ( A cos 4 t + B sin 4 t ) have ‘ x = ’. 15 − 14 t 15 15 1 − 14 t 15 15 1 − t *M1 Using product rule. − 9 e x ' = 4 e ( − A sin 4 t + B cos 4 t ) − 4 e ( A cos 4 t + B sin 4 t ) 1 15 1 1 1 1 DM1 A1 Substitutes initial conditions and forms simultaneous 0 = A + 9 0 = 4 B − 4 A − 9  A = − 9 , B = 3 15 equations. 1 − 14 t 1 15 1 15 1 − t A1 Must have ‘ x = ’. + 9 e x = 3 e ( − 3 cos 4 t + 15 sin 4 t ) 10

This question in 9231/22 May/June 2025

7 Find the solution of the differential equation dy x + 5 - y = 1, dx x 2 + 10x + 61 given that y = 0 when x = 3 . Give your answer in an exact form. [10] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: Finds integrating factor using dx = ln f ( x ) . x + 10 x + 61 dx = 2 ln ( )  f ( x )7  x x + 5 1 2 M1 A1 f '( x ) 2 + 10 x + 61 − 12 ln ( x 2 +10 x + 61) 1 A1 e = x 2 + 10 x + 61 d   y = 1 M1 Correct form on LHS using their integrating factor.  2  2 dx  x + 10 x + 61  x + 10 x + 61 2 2 B1 Completes the square. x + 10 x + 61 = ( x + 5 ) + 36  1 −1  x + 5 *M1 A1 Integrates RHS. Attempting to integrate by parts is dx = sinh + C M0.    6  2 + 6 2  ( x + 5 ) 8 8 C = − sinh −1 ( 6 ) = − ln ( 6 + 53 ) = − ln3 DM1 A1 Substitutes initial conditions.  y −1  x + 5  −1 4 A1 y  x + 5 x 2 + 10 x + 61 = sinh   − sinh ( 3 ) Accept = ln  +  . 2  x + 10 x + 61  6  18 18  x 2 + 10 x + 61   10

This question in 9231/22 May/June 2025

4 Find the particular solution of the differential equation d 2 x d x 2 + - 2x = 2t + t - 1, dt 2 d t dx given that, when t = 0 , x = = 0 . [10] dt … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 4 m2 + m − 2 = 0  m = 1, −2 M1 Auxiliary equation. M1 for correct type of roots. x = Aet + Be−2t A1 Complementary function. Allow “ x = ” missing. Using x instead of t is A0. x = pt 2 + qt + r  x ' = 2 pt + q  x '' = 2 p B1 Particular integral and its derivatives. p = −1 2 p − 2 q = 1 2 p + q − 2 r = −1 M1 Substitutes and equates coefficients. p = −1 q = − 32 r = − 54 A1 x = A e t + B e− 2 t − t 2 − 32 t − 54 A1 FT General solution. FT on CF and their values of p, q and r. Must have “ x = ”. x ' = Ae t − 2 Be −2 t − 2t − 32 *M1 Differentiates. A + B − 54 = 0 A − 2 B − 32 = 0  A = 43 , B = − 121 DM1 A1 Uses initial conditions. x = 43 e t − 121 e−2 t − t 2 − 23 t − 54 A1 Must have “ x = ”. 10

This question in 9231/23 May/June 2025

7 Find the solution of the differential equation dy 2x + 6 - y = 4 , dx x 2 + 6x + 5 given that y = 0 when x = 0 . Give your answer in an exact form. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7 2 x + 6 2 M1 A1 Finds integrating factor using dx = ln x + 6 x + 5  x 2 + 6 x + 5 ( ) f '( x ) dx = ln f( x ) .  f( x ) − ln ( x 2 + 6 x + 5 ) 1 A1 e = x 2 + 6 x + 5 d   2 y = 2 4 M1 Correct form on LHS. dx  x + 6 x + 5  x + 6 x + 5 4 1 1 B1 Partial fractions. Or completing the square. = − x 2 + 6 x + 5 x + 1 x + 5 4  x + 1  *M1 A1 Integrates RHS. (RHS must be of the correct dx = ln   + C 2 form.)  x + 6 x + 5  x + 5  0 = ln 15 + C DM1 Substitutes initial conditions. y  x + 1  1  5 x + 5  A1 2  5 x + 5  x + 6 x + 5 ln   . 2 = ln   − ln 5 = ln   Accept y = ( ) x + 6 x + 5  x + 5   x + 5   x + 5  9

This question in 9231/23 May/June 2025