Second order differential equations — WFM02 IAL Further Mathematics practice questions · Study-IAL

Second order differential equations

WFM02Topic 2.53 questions

Linear equations with constant coefficients using complementary functions and particular integrals.

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Question 17 marks
(a) Find the general solution of $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - 3y = 6x - 1$

(b) Find the particular solution for which $y = 0$ and $\dfrac{dy}{dx} = 0$ when $x = 0$ .
Mark scheme
- M1 Auxiliary equation $m^2 + 2m - 3 = 0 \Rightarrow m = 1, -3$ (1)
- Complementary function (1)
Question 24 marks
Find the general solution of $\frac{d^2y}{dx^2} + 4y = 8\sin 2x$

x = 0

y = Ae^{x} + Be^{-3x}

M1 Tries

y = px + q

and substitutes:

2p - 3(px + q) = 6x - 1

(1)

p = -2

q = -1

, so

y = Ae^{x} + Be^{-3x} - 2x - 1

(1)

M1 Uses

y(0) = 0

A + B - 1 = 0

(1)

M1 Differentiates and uses

y'(0) = 0

A - 3B - 2 = 0

(1)

A = \tfrac{5}{4}

B = -\tfrac{1}{4}

y = \tfrac{5}{4}e^{x} - \tfrac{1}{4}e^{-3x} - 2x - 1

(1)

4y=

8sin2x

Mark scheme

B1 Complementary function $y = A\cos 2x + B\sin 2x$ (1)
B1 Since $\sin 2x$ is in the CF, tries $y = \lambda x\cos 2x + \mu x\sin 2x$ (1)
M1 Differentiates twice and substitutes: $-4\lambda\sin 2x + 4\mu\cos 2x = 8\sin 2x$ (1)
A1 $\lambda = -2$ , $\mu = 0$ : $y = A\cos 2x + B\sin 2x - 2x\cos 2x$ (1)

e−3x

−

2x−

Bsin2x−

2xcos2x