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A-Level · Core Pure Mathematics (Further Maths)

FP.CP8 Differential Equations

EDEXCEL 9FM0 · Calculator allowed · about 165 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A curve satisfies the first order linear differential equation dy/dx + (2/x)y = x², for x > 0.

(a)Show that the integrating factor for this differential equation is x^2.(2)

(b)Hence find the general solution of the differential equation, giving y in terms of x.(4)

(c)Given that y = 3/2 when x = 1, find the particular solution of the differential equation.(3)

(Total for Question 1 is 9 marks)

In an electrical circuit, the charge Q coulombs on a capacitor at time t seconds satisfies the differential equation dQ/dt + 5Q = 15, with Q = 0 when t = 0.

(a)Show that the integrating factor for this differential equation is e^5t.(2)

(b)Hence show that Q = 3(1 - e^-5t).(4)

(c)State the maximum charge that the capacitor can hold, explaining your reasoning.(2)

(d)Find, to 3 significant figures, the time taken for the charge to reach 2.9 coulombs.(3)

(Total for Question 2 is 11 marks)

A curve satisfies the second order differential equation d^2y/dx² - 5(dy/dx) + 6y = 0.

(a)Write down the auxiliary equation and hence find its roots.(2)

(b)Write down the general solution of the differential equation.(2)

(c)Given that y = 1 and dy/dx = 4 when x = 0, find the particular solution of the differential equation.(4)

(Total for Question 3 is 8 marks)

A curve satisfies the second order differential equation d^2y/dx² - 6(dy/dx) + 9y = 0.

(a)Show that the auxiliary equation has a repeated root, and state its value.(2)

(b)Write down the general solution of the differential equation.(1)

(c)Given that y = 2 and dy/dx = 1 when x = 0, find the particular solution of the differential equation.(5)

(Total for Question 4 is 8 marks)

A curve satisfies the second order differential equation d^2y/dx² + 4(dy/dx) + 13y = 0.

(a)Show that the roots of the auxiliary equation are -2 + 3i and -2 - 3i.(3)

(b)State the general solution of the differential equation.(2)

(c)Given that y = 0 and dy/dx = 6 when x = 0, find the particular solution of the differential equation.(4)

(d)Describe the long-term behaviour of y as x tends to infinity, giving a reason for your answer.(2)

(Total for Question 5 is 11 marks)

It is given that y = e^-x(3cos2x - sin2x).

(a)Show that y satisfies the differential equation d^2y/dx² + 2(dy/dx) + 5y = 0.(5)

(b)State the general solution of the differential equation d^2y/dx² + 2(dy/dx) + 5y = 0, justifying the form of your answer by reference to the auxiliary equation.(3)

(c)Find, to 3 significant figures, the two values of x in the interval 0 ≤ x ≤ π for which y = 0, where y is the function given at the start of the question.(4)

(Total for Question 6 is 12 marks)

A curve satisfies the second order differential equation d^2y/dx² - 3(dy/dx) - 4y = 8x² - 2.

(a)Find the complementary function of the differential equation.(3)

(b)By trying a particular integral of the form y = ax² + bx + c, find a particular integral for the differential equation.(5)

(c)Hence write down the general solution of the differential equation.(1)

(d)Given that y = 0 and dy/dx = 0 when x = 0, find the particular solution of the differential equation.(4)

(Total for Question 7 is 13 marks)

A curve satisfies the second order differential equation d^2y/dx² - 4(dy/dx) + 3y = e^3x.

(a)Show that m = 1 and m = 3 are roots of the auxiliary equation, hence state the complementary function.(3)

(b)Explain why a particular integral of the form y = Ce^3x is not suitable for this differential equation, and state an appropriate alternative form to try.(2)

(c)Using your alternative form from part (b), find the particular integral.(5)

(d)Write down the general solution of the differential equation.(1)

(Total for Question 8 is 11 marks)

A curve satisfies the second order differential equation d^2y/dx² + y = sin2x.

(a)Find the complementary function of the differential equation.(2)

(b)By trying a particular integral of the form y = P cos2x + Q sin2x, find the values of P and Q.(5)

(c)Hence state the general solution of the differential equation.(1)

(d)Given that y = 1 and dy/dx = 0 when x = 0, find the particular solution of the differential equation.(5)

(Total for Question 9 is 13 marks)

A particle of mass 0.5 kg is attached to a spring and moves through a resistive medium. The displacement x metres from equilibrium at time t seconds satisfies the differential equation 0.5(d^2x/dt²) + 2(dx/dt) + 10x = 0.

(a)Show that the differential equation can be written as d^2x/dt² + 4(dx/dt) + 20x = 0.(1)

(b)Find the general solution for x in terms of t.(4)

(c)The particle is released from rest at a displacement of 0.5 m from equilibrium. Find x in terms of t.(4)

(d)State, with a reason, what happens to the displacement of the particle as t becomes large.(2)

(e)Determine whether the motion described is oscillatory, critically damped or overdamped, justifying your answer with reference to the roots of the auxiliary equation.(2)

(Total for Question 10 is 13 marks)

The differential equation x²(d^2y/dx²) + x(dy/dx) - 4y = 0, for x > 0, is to be solved using the substitution x = e^u. You may use, without proof, that when x = e^u, x(dy/dx) = dy/du and x²(d^2y/dx²) = (d^2y/du²) - (dy/du).

(a)Using the given substitution results, show that the differential equation transforms to d^2y/du² - 4y = 0.(3)

(b)Hence find y in terms of u, and show that the general solution can be written as y = Ax² + B/x^2.(4)

(c)Given that y = 3 and dy/dx = -1 when x = 1, find the particular solution of the differential equation.(5)

(Total for Question 11 is 12 marks)

Two variables x and y satisfy the simultaneous first order differential equations dx/dt = x + 2y and dy/dt = 3x + 2y. Given that x = 5 and y = 0 when t = 0, answer the following.

(a)By eliminating y, show that x satisfies the differential equation d^2x/dt² - 3(dx/dt) - 4x = 0.(4)

(b)Hence find x in terms of t, given that x = 5 and y = 0 when t = 0.(5)

(c)Find y in terms of t.(3)

(Total for Question 12 is 12 marks)

A forced oscillator satisfies the differential equation d^2x/dt² + 6(dx/dt) + 25x = 51 cos t, where x is displacement in metres and t is time in seconds.

(a)Find the roots of the auxiliary equation, and hence state the complementary function.(3)

(b)By trying a particular integral of the form x = P cos t + Q sin t, show that P = 2 and Q = 0.5.(5)

(c)Given that x = 0 and dx/dt = 0 when t = 0, find the particular solution of the differential equation.(5)

(d)State what is meant by the transient and steady state parts of the solution, identifying each part in your answer to part (c).(3)

(Total for Question 13 is 16 marks)

Mark scheme · FP.CP8 Differential Equations

Question 1

(a) M1 attempts integrating factor e^{integral of 2/x dx} = e^{2 ln x}
(a) A1 correctly simplifies e^{2 ln x} = e^{ln x²} = x², cso (answer printed)
(a) Answer: Integrating factor = x²
(b) M1 multiplies through by x² to obtain d/dx(x² y) = x² . x²
(b) A1 correct right hand side x⁴
(b) M1 integrates both sides: x² y = (x⁵)/5 + C
(b) A1 y = (x³)/5 + C/x², oe
(b) Answer: y = x³/5 + C/x²
(c) M1 substitutes x = 1, y = 3/2 into the general solution
(c) A1 C = 13/10, oe
(c) A1 y = x³/5 + 13/(10x²), cao
(c) Answer: y = x³/5 + 13/(10x²)

Question 2

(a) M1 attempts integrating factor e^{integral of 5 dt}
(a) A1 IF = e^5t, cso (answer printed)
(a) Answer: Integrating factor = e^5t
(b) M1 multiplies through by e^5t and integrates: Q e^5t = integral of 15 e^5t dt
(b) A1 Q e^5t = 3 e^5t + c, oe
(b) M1 uses Q = 0 when t = 0 to find c = -3
(b) A1 Q = 3 - 3e^-5t = 3(1 - e^-5t), cso (answer printed)
(b) Answer: Q = 3(1 - e^-5t)
(c) B1 maximum charge = 3 coulombs
(c) B1 correct reason: as t tends to infinity, e^-5t tends to 0, so Q tends to 3
(c) Answer: Maximum charge = 3 coulombs
(d) M1 sets up 2.9 = 3(1-e^-5t)
(d) M1 rearranges to e^-5t = 1/30 and takes natural logs, t = -(1/5)ln(1/30)
(d) A1 t = 0.680 s (3sf), awrt
(d) Answer: t = 0.680 s (3sf)

Question 3

(a) M1 auxiliary equation m² - 5m + 6 = 0 factorised as (m-2)(m-3) = 0
(a) A1 m = 2, m = 3, cao (both required)
(a) Answer: m = 2, m = 3
(b) B1 correct form y = Ae^2x + Be^3x
(b) B1 correct use of two arbitrary constants A and B, ft from part (a)
(b) Answer: y = Ae^2x + Be^3x
(c) M1 differentiates general solution: dy/dx = 2Ae^2x + 3Be^3x
(c) M1 forms two simultaneous equations A+B=1 and 2A+3B=4 using the given conditions
(c) A1 A = -1, B = 2 (both required)
(c) A1 y = -e^2x + 2e^3x, cao
(c) Answer: y = -e^2x + 2e^3x

Question 4

(a) M1 auxiliary equation m²-6m+9=0 written as (m-3)²=0, or discriminant = 36-36 = 0 shown
(a) A1 repeated root m = 3, cso
(a) Answer: Repeated root m = 3
(b) B1 y = (A + Bx)e^3x, ft from part (a)
(b) Answer: y = (A + Bx)e^3x
(c) M1 differentiates using the product rule
(c) A1 correct dy/dx = (3A+B+3Bx)e^3x
(c) M1 substitutes x=0 into y and dy/dx to form equations A=2 and 3A+B=1
(c) A1 A = 2, B = -5 (both required)
(c) A1 y = (2-5x)e^3x, cao
(c) Answer: y = (2 - 5x)e^3x

Question 5

(a) M1 writes auxiliary equation m²+4m+13=0
(a) M1 uses the quadratic formula or completes the square: m = (-4 ± √16-52)/2
(a) A1 m = -2 ± 3i, cso (both roots required, answer printed)
(a) Answer: m = -2 + 3i or m = -2 - 3i
(b) M1 recognises complex roots p ± qi give solutions of the form e^px(A cos(qx) + B sin(qx))
(b) A1 y = e^-2x(A cos3x + B sin3x), cao
(b) Answer: y = e^-2x(A cos3x + B sin3x)
(c) M1 differentiates the general solution using the product rule
(c) M1 substitutes x=0 into y and dy/dx to form equations A=0 and -2A+3B=6
(c) A1 A = 0, B = 2 (both required)
(c) A1 y = 2e^-2x sin3x, cao
(c) Answer: y = 2e^-2x sin3x
(d) B1 y oscillates with decreasing amplitude, tending to 0 as x tends to infinity
(d) B1 correct reason: e^-2x tends to 0 while sin3x remains bounded between -1 and 1
(d) Answer: y tends to 0 as x tends to infinity (a damped oscillation)

Question 6

(a) M1 differentiates y using the product rule
(a) A1 correct dy/dx = -5e^-x(cos2x + sin2x)
(a) M1 differentiates again using the product rule
(a) A1 correct d^2y/dx² = e^-x(-5cos2x + 15sin2x)
(a) A1 substitutes into d^2y/dx² + 2(dy/dx) + 5y and shows the cos2x and sin2x coefficients both sum to 0, cso
(a) Answer: d^2y/dx² + 2(dy/dx) + 5y = 0, as required
(b) M1 solves auxiliary equation m²+2m+5=0 using the quadratic formula
(b) A1 m = -1 ± 2i
(b) B1 general solution y = e^-x(C cos2x + D sin2x), oe
(b) Answer: y = e^-x(C cos2x + D sin2x)
(c) M1 sets 3cos2x - sin2x = 0 and rearranges to tan2x = 3
(c) M1 finds two values of 2x in [0, 2pi] satisfying tan2x = 3
(c) A1 x = 0.625 (3sf)
(c) A1 x = 2.20 (3sf)
(c) Answer: x = 0.625 or x = 2.20 (3sf)

Question 7

(a) M1 auxiliary equation m²-3m-4=0
(a) A1 factorises to (m-4)(m+1)=0, giving m=4, m=-1
(a) A1 complementary function y = Ae^4x + Be^-x, ft
(a) Answer: CF: y = Ae^4x + Be^-x
(b) M1 differentiates trial twice: dy/dx=2ax+b, d^2y/dx²=2a
(b) M1 substitutes into the differential equation and compares coefficients of x², x and constants
(b) A1 a = -2
(b) A1 b = 3
(b) A1 c = -11/4
(b) Answer: PI: y = -2x² + 3x - 11/4
(c) B1 y = Ae^4x + Be^-x - 2x² + 3x - 11/4, ft
(c) Answer: y = Ae^4x + Be^-x - 2x² + 3x - 11/4
(d) M1 substitutes x=0, y=0 into the general solution to form A+B-11/4=0
(d) M1 differentiates and substitutes x=0, dy/dx=0 to form 4A-B+3=0
(d) A1 A = -1/20 (or B = 14/5)
(d) A1 A = -1/20 and B = 14/5, both stated in y = -1/20 e^4x + 14/5 e^-x - 2x² + 3x - 11/4, cao
(d) Answer: y = -1/20 e^4x + 14/5 e^-x - 2x² + 3x - 11/4

Question 8

(a) M1 auxiliary equation m²-4m+3=0
(a) A1 factorises to (m-1)(m-3)=0, giving m=1, m=3, cso
(a) B1 complementary function y = Ae^x + Be^3x
(a) Answer: CF: y = Ae^x + Be^3x
(b) B1 e^3x is already part of the complementary function (a solution of the homogeneous equation), so the standard trial would give zero and cannot match the right hand side
(b) B1 states the correct alternative trial y = Cxe^3x
(b) Answer: Try y = Cxe^3x
(c) M1 differentiates y=Cxe^3x using the product rule to find dy/dx
(c) A1 correct dy/dx = Ce^3x(1+3x) and d^2y/dx² = Ce^3x(6+9x)
(c) M1 substitutes into the differential equation and simplifies
(c) A1 obtains 2Ce^3x = e^3x
(c) A1 C = 1/2, cao
(c) Answer: PI: y = (1/2)xe^3x
(d) B1 y = Ae^x + Be^3x + (1/2)xe^3x, ft
(d) Answer: y = Ae^x + Be^3x + (1/2)xe^3x

Question 9

(a) B1 auxiliary equation m²+1=0 gives m = +-i
(a) B1 complementary function y = A cosx + B sinx
(a) Answer: CF: y = A cosx + B sinx
(b) M1 differentiates trial twice: d^2y/dx² = -4Pcos2x - 4Qsin2x
(b) M1 substitutes into d^2y/dx²+y and compares coefficients: -3Pcos2x-3Qsin2x = sin2x
(b) A1 P = 0
(b) A1 Q = -1/3
(b) A1 particular integral y = -(1/3)sin2x correctly stated
(b) Answer: P = 0, Q = -1/3
(c) B1 y = A cosx + B sinx - (1/3)sin2x, ft
(c) Answer: y = A cosx + B sinx - (1/3)sin2x
(d) M1 differentiates the general solution: dy/dx = -Asinx+Bcosx-(2/3)cos2x
(d) A1 correct derivative
(d) M1 substitutes x=0, y=1 to obtain A = 1
(d) M1 substitutes x=0, dy/dx=0 to obtain B = 2/3
(d) A1 y = cosx + (2/3)sinx - (1/3)sin2x, cao
(d) Answer: y = cosx + (2/3)sinx - (1/3)sin2x

Question 10

(a) B1 multiplies through by 2 to obtain d^2x/dt²+4(dx/dt)+20x=0, cso (answer printed)
(a) Answer: d^2x/dt² + 4(dx/dt) + 20x = 0
(b) M1 auxiliary equation m²+4m+20=0
(b) M1 uses the quadratic formula: m = (-4 ± √16-80)/2
(b) A1 m = -2 ± 4i
(b) A1 x = e^-2t(A cos4t + B sin4t)
(b) Answer: x = e^-2t(A cos4t + B sin4t)
(c) M1 uses x(0)=0.5 to obtain A=0.5
(c) M1 differentiates and uses dx/dt=0 at t=0 to form -2A+4B=0
(c) A1 B = 0.25
(c) A1 x = e^-2t(0.5cos4t + 0.25sin4t), cao
(c) Answer: x = e^-2t(0.5 cos4t + 0.25 sin4t)
(d) B1 x tends to 0 as t becomes large
(d) B1 correct reason: e^-2t tends to 0 while the trigonometric factor remains bounded, so the oscillation dies away
(d) Answer: x tends to 0 as t becomes large
(e) B1 the motion is (damped) oscillatory / underdamped
(e) B1 correct justification: the auxiliary equation has complex (non-real) roots -2+-4i, so the solution is a decaying exponential multiplying a genuine oscillation, rather than a single or double real root as in critical or overdamping
(e) Answer: The motion is damped oscillatory (underdamped)

Question 11

(a) M1 substitutes the given expressions for x²(d^2y/dx²) and x(dy/dx) into the differential equation
(a) A1 obtains (d^2y/du² - dy/du) + (dy/du) - 4y = 0 with the -dy/du and +dy/du terms identified
(a) A1 correctly cancels to give d^2y/du² - 4y = 0, cso (answer printed)
(a) Answer: d^2y/du² - 4y = 0
(b) M1 auxiliary equation m²-4=0 solved to give m = ±2
(b) A1 y = Ae^2u + Be^-2u
(b) M1 uses u = ln x to write e^2u = x² and e^-2u = x^-2
(b) A1 y = Ax² + B/x², cso
(b) Answer: y = Ax² + B/x²
(c) M1 differentiates y=Ax²+B/x² to obtain dy/dx=2Ax-2B/x³
(c) M1 substitutes x=1 into y and dy/dx to form A+B=3 and 2A-2B=-1
(c) A1 A = 5/4 (or 1.25)
(c) A1 B = 7/4 (or 1.75)
(c) A1 y = (5/4)x² + (7/4)/x², cao
(c) Answer: y = (5/4)x² + (7/4)/x²

Question 12

(a) M1 rearranges dx/dt = x+2y to express y in terms of x and dx/dt: y = (dx/dt - x)/2
(a) M1 differentiates dx/dt = x+2y and substitutes dy/dt = 3x+2y to eliminate dy/dt
(a) A1 substitutes the expression for y (or 4y) to eliminate the remaining y term
(a) A1 obtains d^2x/dt² - 3(dx/dt) - 4x = 0, cso (answer printed)
(a) Answer: d^2x/dt² - 3(dx/dt) - 4x = 0
(b) M1 auxiliary equation m²-3m-4=0 solved to give m=4, m=-1
(b) A1 x = Ae^4t + Be^-t
(b) M1 uses x(0)=5 and y(0)=0 (via y=(dx/dt-x)/2 from part a) to form simultaneous equations in A and B
(b) A1 A = 2, B = 3 (both required)
(b) A1 x = 2e^4t + 3e^-t, cao
(b) Answer: x = 2e^4t + 3e^-t
(c) M1 differentiates x(t) found in part (b): dx/dt = 8e^4t - 3e^-t
(c) M1 substitutes into y = (dx/dt - x)/2 from part (a)
(c) A1 y = 3e^4t - 3e^-t, cao
(c) Answer: y = 3e^4t - 3e^-t

Question 13

(a) M1 auxiliary equation m²+6m+25=0
(a) M1 uses the quadratic formula: m = (-6 ± √36-100)/2
(a) A1 m = -3 ± 4i, complementary function x = e^-3t(Acos4t+Bsin4t), cso
(a) Answer: CF: x = e^-3t(A cos4t + B sin4t)
(b) M1 differentiates trial twice: d^2x/dt² = -Pcost - Qsint
(b) M1 substitutes into the differential equation and compares coefficients of cos t and sin t
(b) A1 forms the pair of equations 24P+6Q=51 and -6P+24Q=0, oe
(b) A1 P = 2, cso
(b) A1 Q = 0.5, cso
(b) Answer: P = 2, Q = 0.5
(c) M1 differentiates the general solution x=e^-3t(Acos4t+Bsin4t)+2cost+0.5sint
(c) M1 substitutes t=0, x=0 to obtain A+2=0, so A=-2
(c) M1 substitutes t=0, dx/dt=0 to obtain -3A+4B+0.5=0
(c) A1 B = -13/8 (or -1.625)
(c) A1 x = e^-3t(-2cos4t - 1.625sin4t) + 2cost + 0.5sint, cao
(c) Answer: x = e^-3t(-2cos4t - 1.625sin4t) + 2cost + 0.5sint
(d) B1 the transient part is e^-3t(-2cos4t - 1.625sin4t), which decays to zero as t becomes large because of the e^-3t factor
(d) B1 the steady state part is 2cost + 0.5sint, which persists indefinitely, oscillating with the same period as the forcing term
(d) B1 as t becomes large, x(t) approaches the steady state part 2cost + 0.5sint since the transient part dies away
(d) Answer: Transient: e^-3t(-2cos4t-1.625sin4t); Steady state: 2cost+0.5sint

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