Second-order differential equation (particular solution)

Differential equations

A-level Further Maths (9FM0) · Core Pure

ORIGINAL

Auxiliary equation, complementary function, particular integral, then the boundary conditions — in that order.

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Question

Find the particular solution of

\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + 2y = 4x

for which

y = 0

and

\dfrac{dy}{dx} = 1

when

x = 0

Every step worked, with the reasoning.

1
Auxiliary equation: $m^2 - 3m + 2 = 0 \Rightarrow (m - 1)(m - 2) = 0 \Rightarrow m = 1,\ 2$
Replace $\frac{d^2y}{dx^2}$ , $\frac{dy}{dx}$ , $y$ by $m^2$ , $m$ , $1$ in the homogeneous equation.
2
Complementary function: $y = Ae^{x} + Be^{2x}$
Two distinct real roots $m_1, m_2$ give $Ae^{m_1x} + Be^{m_2x}$ .
3
Try $y = px + q$ : $\dfrac{dy}{dx} = p$ , $\dfrac{d^2y}{dx^2} = 0$ , so $0 - 3p + 2(px + q) = 4x$
The right-hand side is linear, so the particular integral is a general linear function.
4
$x$ : $2p = 4 \Rightarrow p = 2$ ; constants: $2q - 3p = 0 \Rightarrow q = 3$
Compare coefficients of $x$ and the constant terms.
5
General solution: $y = Ae^{x} + Be^{2x} + 2x + 3$
General solution $=$ complementary function $+$ particular integral.
6
$x = 0,\ y = 0$ : $A + B + 3 = 0$ . $\ \dfrac{dy}{dx} = Ae^{x} + 2Be^{2x} + 2$ , so $x = 0$ : $A + 2B + 2 = 1$
Apply both conditions — differentiate the general solution for the gradient condition.
7
$A + B = -3$ and $A + 2B = -1 \Rightarrow B = 2,\ A = -5$
Subtract the equations to find $B$ , then substitute back.

Answer: $y = 2e^{2x} - 5e^{x} + 2x + 3$