Question:
Particular solution of $y'' + y = \cos x$ is
Particular solution of $y'' + y = \cos x$ is
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When $f(D) = D^2 + a^2$ and the input is $\cos ax$ or $\sin ax$, always multiply by $x$ and differentiate the denominator.
Updated On: Jul 3, 2026
- $\frac{x}{2} \sin x$
- $\frac{1}{2} \sin x$
- $\sin x$
- $x \cos x$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The particular integral (P.I.) is given by $\frac{1}{D^{2} + 1} \cos x$. Since putting $D^{2} = -a^{2} = -1$ makes the denominator zero, this is a case of failure.
Step 2: Meaning
The general formula for $\frac{1}{D^{2} + a^{2}} \cos ax$ when it fails is $\frac{x}{2a} \sin ax$.
Step 3: Analysis
Here $a = 1$. Applying the formula: $P.I. = \frac{x}{2(1)} \sin(1)x = \frac{x}{2} \sin x$.
Step 4: Conclusion
Therefore, the particular solution specifically addressing the non-homogeneous part is $\frac{x}{2} \sin x$. Final Answer: (A)
The particular integral (P.I.) is given by $\frac{1}{D^{2} + 1} \cos x$. Since putting $D^{2} = -a^{2} = -1$ makes the denominator zero, this is a case of failure.
Step 2: Meaning
The general formula for $\frac{1}{D^{2} + a^{2}} \cos ax$ when it fails is $\frac{x}{2a} \sin ax$.
Step 3: Analysis
Here $a = 1$. Applying the formula: $P.I. = \frac{x}{2(1)} \sin(1)x = \frac{x}{2} \sin x$.
Step 4: Conclusion
Therefore, the particular solution specifically addressing the non-homogeneous part is $\frac{x}{2} \sin x$. Final Answer: (A)
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