UPSC 2018 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 13 yrs · Read the full method →

Question

Solve: $y''-y=x^2e^{2x}$ .

CF from auxiliary equation; PI by exponential shift $\frac1{\phi(D)}e^{2x}u=e^{2x}\frac1{\phi(D+2)}u$ then operator series to order $D^2$ .

This is a constant-coefficient linear ODE; write it as $(D^2-1)y=x^2e^{2x}$ where $D=\dfrac{d}{dx}$ .

Auxiliary equation $m^2-1=0\Rightarrow m=\pm1$ , so

y_c=C_1e^{x}+C_2e^{-x}.

y_p=\frac{1}{D^2-1}\,x^2e^{2x}.

Since $2$ is not a root of $m^2-1$ , the forcing is non-resonant. Use the shift rule $\dfrac{1}{\phi(D)}e^{2x}u=e^{2x}\dfrac{1}{\phi(D+2)}u$ :

y_p=e^{2x}\,\frac{1}{(D+2)^2-1}\,x^2=e^{2x}\,\frac{1}{D^2+4D+3}\,x^2.

Factor out $3$ and expand $\dfrac{1}{1+\frac{4D+D^2}{3}}$ to order $D^2$ (higher powers annihilate $x^2$ ):

\frac{1}{3}\Big(1+\tfrac{4D+D^2}{3}\Big)^{-1} =\frac13\Big[1-\tfrac{4D+D^2}{3}+\big(\tfrac{4D}{3}\big)^2-\cdots\Big] =\frac13\Big[1-\tfrac{4}{3}D+\big(\tfrac{16}{9}-\tfrac13\big)D^2\Big].

The $D^2$ coefficient is $\dfrac{16}{9}-\dfrac13=\dfrac{16-3}{9}=\dfrac{13}{9}$ . Apply to $x^2$ (with $Dx^2=2x,\ D^2x^2=2$ ):

\frac1{D^2+4D+3}x^2=\frac13\Big[x^2-\tfrac43(2x)+\tfrac{13}{9}(2)\Big] =\frac13\Big[x^2-\tfrac83x+\tfrac{26}{9}\Big]=\frac{9x^2-24x+26}{27}.

Hence

y_p=\frac{(9x^2-24x+26)}{27}\,e^{2x}.

\boxed{\;y=C_1e^{x}+C_2e^{-x}+\frac{1}{27}\big(9x^2-24x+26\big)e^{2x}.\;}