UPSC Maths 2025 Paper 1 Q8a — Solution

15 marks · Section B

Question

Solve the differential equation $(x + 2)\dfrac{d^2 y}{dx^2} - (2x + 5)\dfrac{dy}{dx} + 2y = (1 + x)e^x$ by the method of variation of parameters.

Technique

For this variable-coefficient linear ODE, first find two independent complementary solutions (by inspecting exponential and polynomial trial forms), then apply the variation-of-parameters formula $y_p = -y_1\!\int\!\frac{y_2 R}{W}\,dx + y_2\!\int\!\frac{y_1 R}{W}\,dx,$ where $R$ is the RHS after writing the ODE in standard form $y'' + Py' + Qy = R$ .

Solution

Complementary functions

Trial $y = e^{mx}$ in the homogeneous equation $(x+2)y'' - (2x+5)y' + 2y = 0$ : $(x+2)m^2 - (2x+5)m + 2 = 0 \text{ for all } x.$ Coefficient of $x$ : $m^2 - 2m = 0 \Rightarrow m = 0$ or $m = 2$ . Constant term: $2m^2 - 5m + 2 = 0 \Rightarrow m = 2$ or $m=\tfrac12$ . The common value is $m=2$ , giving $y_1 = e^{2x}.$

For a second solution try a polynomial $y = \alpha x + \beta$ : $y''=0$ , $y'=\alpha$ , so $-(2x+5)\alpha + 2(\alpha x+\beta) = -2\alpha x - 5\alpha + 2\alpha x + 2\beta = 2\beta - 5\alpha = 0$ , giving $\beta=\tfrac52\alpha$ . Take $\alpha=2$ : $y_2 = 2x + 5.$ (Direct substitution confirms both annihilate the homogeneous operator.)

Standard form and Wronskian

Divide by $(x+2)$ : $y'' + P y' + Q y = R$ with $R = \dfrac{(1+x)e^x}{x+2}$ .

Wronskian: $W = y_1 y_2' - y_2 y_1' = e^{2x}\cdot 2 - (2x+5)\cdot 2e^{2x} = 2e^{2x}\bigl[1 - (2x+5)\bigr] = -4(x+2)e^{2x}.$

Variation of parameters

$y_p = -y_1\int\frac{y_2 R}{W}\,dx + y_2\int\frac{y_1 R}{W}\,dx.$

Compute the integrands ( $R/W = \dfrac{(1+x)e^x}{(x+2)\cdot(-4)(x+2)e^{2x}} = -\dfrac{(1+x)e^{-x}}{4(x+2)^2}$ ):

$u_2' = \frac{y_1 R}{W} = \frac{e^{2x}(1+x)e^x}{-4(x+2)e^{2x}(x+2)} = -\frac{(x+1)e^{x}}{4(x+2)^2},\qquad u_2 = \int u_2'\,dx = -\frac{e^x}{4(x+2)}.$

(Using $\int\dfrac{(x+1)e^x}{(x+2)^2}dx = \dfrac{e^x}{x+2}$ , since $\dfrac{d}{dx}\dfrac{e^x}{x+2} = \dfrac{e^x(x+2)-e^x}{(x+2)^2} = \dfrac{(x+1)e^x}{(x+2)^2}$ .)

$u_1' = -\frac{y_2 R}{W} = \frac{(x+1)(2x+5)e^{-x}}{4(x+2)^2},\qquad u_1 = \int u_1'\,dx = -\frac{(2x+3)e^{-x}}{4(x+2)}.$

Then $y_p = y_1 u_1 + y_2 u_2 = e^{2x}\!\left(-\frac{(2x+3)e^{-x}}{4(x+2)}\right) + (2x+5)\!\left(-\frac{e^x}{4(x+2)}\right).$ $= -\frac{e^x}{4(x+2)}\Bigl[(2x+3) + (2x+5)\Bigr] = -\frac{e^x}{4(x+2)}\cdot(4x+8) = -\frac{e^x\cdot 4(x+2)}{4(x+2)} = -e^x.$

Complete solution

$y = C_1 e^{2x} + C_2(2x+5) - e^x.$

Answer

$\boxed{\;y = C_1 e^{2x} + C_2\,(2x + 5) - e^x.\;}$