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UPSC 2021 Maths Optional Paper 1 Q8a-ii — Step-by-Step Solution

10 marks · Section B

Method of variation of parameters · ODEs · asked 12× in 14 yrs · Read the full method →

Question

Find general solution of $x^2 y''-2x(1+x)y'+2(1+x)y=0$. Hence solve $x^2 y''-2x(1+x)y'+2(1+x)y=x^3$ by variation of parameters.

Technique

Spot $y_1=x$ by inspection; reduction of order gives $y_2=xe^{2x}$; standard VoP yields particular solution.

Solution

Step 1 — Find one solution by inspection

Try $y=x$: $y'=1$, $y''=0$. Substitute: $x^2\cdot 0-2x(1+x)\cdot 1+2(1+x)\cdot x=-2x(1+x)+2x(1+x)=0$ ✓.

So $y_1=x$.

Step 2 — Find second solution via reduction of order

Try $y_2=v(x)\cdot x$. Then $y_2'=v'x+v$, $y_2''=v''x+2v'$.

Substitute into $x^2 y''-2x(1+x)y'+2(1+x)y=0$: $x^2(v''x+2v')-2x(1+x)(v'x+v)+2(1+x)vx=0$, $x^3 v''+2x^2 v'-2x^2(1+x)v'-2x(1+x)v+2x(1+x)v=0$, $x^3 v''+2x^2 v'-2x^2(1+x)v'=0$.

Simplify $2x^2 v'-2x^2(1+x)v'=2x^2 v'[1-(1+x)]=-2x^3 v'$.

So $x^3 v''-2x^3 v'=0$, i.e., $v''=2v'$.

Let $u=v'$: $u'=2u$, so $u=Ce^{2x}$. Take $C=1$: $v'=e^{2x}$, $v=e^{2x}/2$.

So $y_2=v\cdot x=xe^{2x}/2$. Up to a constant, take $y_2=xe^{2x}$.

Step 3 — Verify $y_2=xe^{2x}$

$y_2'=e^{2x}+2xe^{2x}=(1+2x)e^{2x}$. $y_2''=2e^{2x}+(1+2x)(2)e^{2x}=2e^{2x}+2(1+2x)e^{2x}=(4+4x)e^{2x}=4(1+x)e^{2x}$.

LHS: $x^2\cdot 4(1+x)e^{2x}-2x(1+x)(1+2x)e^{2x}+2(1+x)xe^{2x}$ $=e^{2x}(1+x)[4x^2-2x(1+2x)+2x]$ $=e^{2x}(1+x)[4x^2-2x-4x^2+2x]$ $=e^{2x}(1+x)\cdot 0=0$ ✓.

Step 4 — General homogeneous solution

$y_h=C_1 x+C_2 xe^{2x}$.

Step 5 — Variation of parameters for $x^3$ RHS

Standard form: divide by $x^2$: $y''-\dfrac{2(1+x)}{x}y'+\dfrac{2(1+x)}{x^2}y=x$.

So $g(x)=x$ in standard form $y''+\ldots=g(x)$.

VoP: $y_p=u_1 y_1+u_2 y_2$ with $u_1'y_1+u_2'y_2=0$, $u_1'y_1'+u_2'y_2'=g(x)$.

With $y_1=x$, $y_2=xe^{2x}$, $y_1'=1$, $y_2'=(1+2x)e^{2x}$:

System: $xu_1'+xe^{2x}u_2'=0\Rightarrow u_1'=-e^{2x}u_2'$.

$u_1'+(1+2x)e^{2x}u_2'=x$.

Sub: $-e^{2x}u_2'+(1+2x)e^{2x}u_2'=x$, $e^{2x}u_2'(2x)=x$, $u_2'=\dfrac{1}{2}e^{-2x}$.

$u_2=\int\dfrac{1}{2}e^{-2x}\,dx=-\dfrac{1}{4}e^{-2x}$.

$u_1'=-e^{2x}\cdot\dfrac{1}{2}e^{-2x}=-1/2$.

$u_1=-x/2$.

Step 6 — Particular solution

$y_p=u_1 y_1+u_2 y_2=(-x/2)(x)+(-e^{-2x}/4)(xe^{2x})=-x^2/2-x/4$.

Step 7 — General solution

Answer