UPSC 2019 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 14 yrs · Read the full method →

Question

Determine the complete solution of the differential equation

\frac{d^2y}{dx^2}-4\frac{dy}{dx}+4y=3x^2e^{2x}\sin 2x.

Technique

Exponential shift $y=e^{2x}u$ collapses $(D-2)^2$ to $D^2$ because the forcing carries the resonant factor $e^{2x}$ ; then double integration of $3x^2\sin2x$ .

Solution

Step 1 — Complementary function

The auxiliary equation is $m^2-4m+4=(m-2)^2=0$ , a repeated root $m=2$ . Hence

y_c=(C_1+C_2x)e^{2x}.

Step 2 — Reduce the forcing by the exponential shift

Put $y=e^{2x}u$ . With $D=\dfrac{d}{dx}$ , $(D-2)^2y=4y'...$ more cleanly: $(D-2)\big(e^{2x}u\big)=e^{2x}Du$ , so

(D-2)^2\big(e^{2x}u\big)=e^{2x}D^2u.

The equation $(D-2)^2y=3x^2e^{2x}\sin2x$ becomes

e^{2x}\,u''=3x^2e^{2x}\sin2x\quad\Longrightarrow\quad u''=3x^2\sin2x.

Step 3 — Integrate twice

u'=\int 3x^2\sin2x\,dx.

Using $\int x^2\sin2x\,dx=-\tfrac12x^2\cos2x+\tfrac12 x\sin2x+\tfrac14\cos2x$ ,

u'=3\!\left(-\tfrac12x^2\cos2x+\tfrac12 x\sin2x+\tfrac14\cos2x\right)=-\tfrac32x^2\cos2x+\tfrac32x\sin2x+\tfrac34\cos2x.

Integrate again (dropping constants, which are absorbed into $y_c$ ):

u=-\tfrac34x^2\sin2x-\tfrac32x\cos2x+\tfrac98\sin2x.

(The $x^2\cos2x$ term integrates to $-\tfrac34 x^2\sin2x-\tfrac34 x\cos2x+\ldots$ ; collecting all pieces gives the coefficients above.)

Step 4 — Particular integral and complete solution

y_p=e^{2x}u=e^{2x}\!\left(-\tfrac34x^2\sin2x-\tfrac32x\cos2x+\tfrac98\sin2x\right).

Therefore the complete solution is

Answer

\boxed{\;y=(C_1+C_2x)e^{2x}+e^{2x}\!\left(-\frac34x^2\sin2x-\frac32x\cos2x+\frac98\sin2x\right).\;}