← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q7a — Step-by-Step Solution

15 marks · Section B

Reduction of order with one solution known · ODEs · asked 3× in 13 yrs · Read the full method →

Question

Solve the following differential equation:

when $e^{x}$ is a solution to its corresponding homogeneous differential equation.

Technique

Reduction of order via $y=e^{x}v$; first-order ODE in $w=v'$; integrating factor; recognise antiderivative.

Solution

Strategy. Reduction of order: substitute $y=e^{x}v(x)$ to convert the second-order equation into a first-order ODE in $v'$.

Step 1 — Substitute $y=e^{x}v$

$y'=e^{x}(v+v')$, $y''=e^{x}(v+2v'+v'')$.

LHS:

Collect terms:

• $v''$ coefficient: $x$.
• $v'$ coefficient: $2x-2(x+1)=-2$.
• $v$ coefficient: $x-2(x+1)+(x+2)=x-2x-2+x+2=0$.

So LHS $=e^{x}(xv''-2v')$.

Setting $=$ RHS $=(x-2)e^{2x}$ and dividing by $e^{x}$:

Step 2 — Substitute $w=v'$

This becomes first-order in $w$:

Step 3 — Integrating factor

$\mu=\exp\!\left(-\int\dfrac{2}{x}\,dx\right)=x^{-2}$.

Recognise $\dfrac{d}{dx}\!\left(\dfrac{e^{x}}{x^{2}}\right)=\dfrac{e^{x}}{x^{2}}-\dfrac{2e^{x}}{x^{3}}=\!\left(\dfrac{1}{x^{2}}-\dfrac{2}{x^{3}}\right)e^{x}$.

So $\dfrac{w}{x^{2}}=\dfrac{e^{x}}{x^{2}}+C$, hence $w=e^{x}+Cx^{2}$.

Step 4 — Integrate back

$v'=w=e^{x}+Cx^{2}$. Integrate:

Therefore

Renaming constants ($c_1=D$, $c_2=C/3$):

Answer