← 2020 Paper 1

UPSC 2020 Maths Optional Paper 1 Q8a-i — Step-by-Step Solution

10 marks · Section B

Euler-Cauchy equation · ODEs · asked 8× in 13 yrs · Read the full method →

Question

Solve the following differential equation: $(x+1)^2 y''-4(x+1)y'+6y=6(x+1)^2+\sin\log(x+1)$.

Technique

Cauchy–Euler $\to$ constant-coefficient ODE via $t=\log(x+1)$; operator method for PI, with resonance handling for $6e^{2t}$ and the $\tfrac{1}{1-D}$ trick for $\sin t$.

Solution

This is a Cauchy–Euler (equidimensional) equation in the variable $(x+1)$.

Step 1 — Substitution to constant coefficients

Put

Then the standard operator identities give

The equation becomes

Step 2 — Complementary function

Auxiliary equation $m^2-5m+6=(m-2)(m-3)=0\Rightarrow m=2,3$. Hence

Step 3 — Particular integral for $6e^{2t}$ (resonant case)

Since $m=2$ is a root, $f(2)=0$ and we use $\dfrac{1}{f(D)}e^{2t}=t\,\dfrac{e^{2t}}{f'(2)}$ with $f'(m)=2m-5$, $f'(2)=-1$:

Step 4 — Particular integral for $\sin t$

Replace $D^2\to-1$ in $f(D)=D^2-5D+6$, giving $-1-5D+6=5-5D=5(1-D)$:

(Here $1-D^2$ acting on $\sin t$ gives $1-(-1)=2$.)

Step 5 — Assemble and return to $x$

With $e^t=x+1$ and $t=\log(x+1)$:

Answer