Question:
The particular integral of \(\displaystyle \frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=e^{-2x}\) is
The particular integral of \(\displaystyle \frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=e^{-2x}\) is
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If \(F(a)=0\) while finding P.I. of \(e^{ax}\), use \(\frac{x e^{ax}}{F'(a)}\).
Updated On: May 7, 2026
- \(-xe^{-2x}\)
- \(xe^{-2x}\)
- \(-\frac{x}{2}e^{-2x}\)
- \(\frac{x}{2}e^{-2x}\)
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The Correct Option is A
Solution and Explanation
We are given:
\[
\frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=e^{-2x}.
\]
Using operator notation:
\[
(D^2+3D+2)y=e^{-2x}.
\]
Here,
\[
F(D)=D^2+3D+2.
\]
The particular integral is:
\[
P.I.=\frac{1}{F(D)}e^{-2x}.
\]
For \(e^{ax}\), we use:
\[
F(D)e^{ax}=F(a)e^{ax}.
\]
Here,
\[
a=-2.
\]
Now calculate:
\[
F(-2)=(-2)^2+3(-2)+2.
\]
\[
F(-2)=4-6+2.
\]
\[
F(-2)=0.
\]
Since \(F(-2)=0\), the usual method fails because \(-2\) is a root of the auxiliary equation.
So we use the repeated-root particular integral rule:
\[
P.I.=\frac{x e^{-2x}}{F'(-2)}.
\]
Now find:
\[
F'(D)=2D+3.
\]
Therefore,
\[
F'(-2)=2(-2)+3.
\]
\[
F'(-2)=-4+3=-1.
\]
Hence,
\[
P.I.=\frac{x e^{-2x}}{-1}.
\]
\[
P.I.=-xe^{-2x}.
\]
Therefore, the particular integral is:
\[
-xe^{-2x}.
\]
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