← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q6a — Step-by-Step Solution

15 marks · Section B

Linear ODE with constant coefficients · ODEs · asked 4× in 14 yrs · Read the full method →

Question

Solve the differential equation

Technique

Auxiliary polynomial; resonance modification ($x e^x$) for the $e^x$ piece since $1$ is a root; trigonometric ansatz $B\cos x+C\sin x$ for the $\cos x$ piece (no resonance because $\pm i$ are not roots of the auxiliary).

Solution

Step 1 — Solve the homogeneous part

Auxiliary equation: $m^3-3m^2+4m-2=0$.

Try $m=1$: $1-3+4-2=0$ ✓. So $(m-1)$ is a factor.

Polynomial division: $m^3-3m^2+4m-2=(m-1)(m^2-2m+2)$.

Roots of $m^2-2m+2=0$: $m=1\pm i$.

Complementary function:

Step 2 — Particular integral for $e^x$

Since $m=1$ is a simple root of the auxiliary equation, try $y_{p,1}=A x e^x$.

The operator can be written $D^3-3D^2+4D-2=(D-1)(D^2-2D+2)$.

Apply $(D-1)(D^2-2D+2)$ to $Axe^x$. First $(D^2-2D+2)(Axe^x)$: $D(Axe^x)=Ae^x+Axe^x=Ae^x(1+x).$ $D^2(Axe^x)=Ae^x(1+x)+Ae^x=Ae^x(2+x).$ $(D^2-2D+2)(Axe^x)=Ae^x[(2+x)-2(1+x)+2x]=Ae^x[2+x-2-2x+2x]=Ae^x\cdot x=Axe^x.$

Then $(D-1)(Axe^x)=Ae^x+Axe^x-Axe^x=Ae^x$.

Set this equal to $e^x$: $A=1$, so $y_{p,1}=xe^x$.

Step 3 — Particular integral for $\cos x$

Try $y_{p,2}=B\cos x+C\sin x$.

$y_{p,2}'=-B\sin x+C\cos x.$ $y_{p,2}''=-B\cos x-C\sin x.$ $y_{p,2}'''=B\sin x-C\cos x.$

Substitute: $y_{p,2}'''-3y_{p,2}''+4y_{p,2}'-2y_{p,2}$ $=(B\sin x-C\cos x)-3(-B\cos x-C\sin x)+4(-B\sin x+C\cos x)-2(B\cos x+C\sin x)$ $=\sin x(B+3C-4B-2C)+\cos x(-C+3B+4C-2B)$ $=\sin x(-3B+C)+\cos x(B+3C).$

Set equal to $\cos x$: $-3B+C=0,\,B+3C=1$. From the first: $C=3B$. Substitute: $B+9B=1\Rightarrow B=\tfrac{1}{10},\,C=\tfrac{3}{10}$.

$y_{p,2}=\tfrac{1}{10}\cos x+\tfrac{3}{10}\sin x=\tfrac{\cos x+3\sin x}{10}.$

Step 4 — General solution

Answer