UPSC 2024 Maths Optional Paper 1 Q6c-i — Step-by-Step Solution

10 marks · Section B

Reduction of order with one solution known · ODEs · asked 3× in 14 yrs · Read the full method →

Question

Find the second solution of the differential equation $xy''+(x-1)y'-y=0$ using $u(x)=-e^{-x}$ as one of the solutions.

Reduction of order: set $y=u(x)v(x)$ , derive a first-order ODE for $v'$ , integrate twice.

Step 1 — Substitute $y=u(x)v(x)=-e^{-x}v(x)$ .

y'=e^{-x}(v-v'),\qquad y''=e^{-x}(-v+2v'-v'').

Substitute into the ODE and divide by $e^{-x}$ :

x(-v+2v'-v'')+(x-1)(v-v')+v=0.

Expanding and collecting terms by $v,v',v''$ :

Result: $-xv''+(x+1)v'=0$ .

Step 2 — First-order ODE for $w=v'$ .

xw'=(x+1)w\;\Rightarrow\;\frac{dw}{w}=\left(1+\frac{1}{x}\right)dx\;\Rightarrow\;w=Cxe^x.

Step 3 — Integrate for $v$ .

v=C\int xe^x\,dx=C(x-1)e^x+\text{const}.

Taking $C=1$ and discarding the constant (which re-introduces $u$ ): $v=(x-1)e^x$ .

Step 4 — Second solution.

y_2=u\cdot v=(-e^{-x})(x-1)e^x=-(x-1)=1-x.

Verification. $y_2=1-x$ , $y_2'=-1$ , $y_2''=0$ . Then $xy_2''+(x-1)y_2'-y_2=0+(x-1)(-1)-(1-x)=0$ ✓.

\boxed{\;y_2(x)=1-x.\;}