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UPSC 2022 Maths Optional Paper 1 Q8a-ii — Step-by-Step Solution

10 marks · Section B

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

Question

Solve $(3x+2)^2\dfrac{d^2 y}{dx^2}+5(3x+2)\dfrac{dy}{dx}-3y=x^2+x+1$.

Technique

Substitution $u=3x+2$ converts to Cauchy-Euler form; then $u=e^t$ converts to constant-coefficient; CF from char roots $1/3,-1$; PI for each term in RHS via $1/f(D)$ acting on $e^{at}$.

Solution

Setup. Cauchy-Euler-like equation with $(3x+2)$ playing the role of $x$. Substitute $u=3x+2$ (so $du=3\,dx$, $x=(u-2)/3$).

Step 1 — Convert derivatives

$\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=3\dfrac{dy}{du}$.

$\dfrac{d^2 y}{dx^2}=\dfrac{d}{dx}\!\left(3\dfrac{dy}{du}\right)=3\cdot\dfrac{d^2 y}{du^2}\cdot 3=9\dfrac{d^2 y}{du^2}$.

Substitute: $u^2\cdot 9 y''_u+5u\cdot 3y'_u-3y=x^2+x+1$, $9u^2 y''_u+15u y'_u-3y=x^2+x+1$.

Divide by 3: $3u^2 y''_u+5u y'_u-y=\dfrac{x^2+x+1}{3}$.

Express RHS in terms of $u$: $x=(u-2)/3$, so $x^2=(u-2)^2/9$, $x^2+x+1=(u-2)^2/9+(u-2)/3+1$.

$(u-2)^2/9=(u^2-4u+4)/9$. $(u-2)/3=(3u-6)/9$. $1=9/9$.

Sum: $(u^2-4u+4+3u-6+9)/9=(u^2-u+7)/9$.

RHS becomes $(u^2-u+7)/(9\cdot 3)=(u^2-u+7)/27$.

So

Step 2 — Convert to constant-coefficient ODE via $u=e^t$

Let $u=e^t$, $t=\ln u$. Standard Euler-Cauchy: $u y'=Dy$ where $D=d/dt$. $u^2 y''=D(D-1)y$.

Substitute: $3D(D-1)y+5Dy-y=\dfrac{e^{2t}-e^t+7}{27}$, $[3D^2-3D+5D-1]y=\dfrac{e^{2t}-e^t+7}{27}$, $[3D^2+2D-1]y=\dfrac{e^{2t}-e^t+7}{27}$.

Factor: $3D^2+2D-1=(3D-1)(D+1)$.

Step 3 — Complementary function

Roots of $3D^2+2D-1=0$: $D=(-2\pm\sqrt{4+12})/6=(-2\pm 4)/6=1/3,-1$.

So CF $=C_1 e^{t/3}+C_2 e^{-t}=C_1 u^{1/3}+C_2/u=C_1(3x+2)^{1/3}+C_2/(3x+2)$.

Step 4 — Particular integral

RHS = $\dfrac{1}{27}(e^{2t}-e^t+7)$.

PI for $e^{2t}$: $f(D)=3D^2+2D-1$, $f(2)=12+4-1=15$. PI $=\dfrac{e^{2t}}{15}$.

PI for $-e^t$: $f(1)=3+2-1=4$. PI $=-\dfrac{e^t}{4}$.

PI for $7$ (constant): $f(0)=-1$. PI $=\dfrac{7}{-1}=-7$.

Combine:

Convert back: $e^{2t}=u^2=(3x+2)^2$, $e^t=u=3x+2$.

Step 5 — General solution

Answer