Question:
Particular integral of $y'' + y = \cosh 3x$ is
Particular integral of $y'' + y = \cosh 3x$ is
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For $\cosh ax$ or $\sinh ax$, substitute $D^2 = a^2$. For $\cos ax$ or $\sin ax$, substitute $D^2 = -a^2$.
Updated On: Jul 3, 2026
- $\frac{1}{10} \cosh 3x$
- $\frac{-1}{10} \sinh 3x$
- $3 \cosh x$
- $x \sinh 3x$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
For hyperbolic functions like $\cosh ax$, we substitute $D^{2} = a^{2}$ in the operator.
Step 2: Meaning
$P.I. = \frac{1}{D^{2} + 1} \cosh 3x$. Here $a = 3$, so $a^{2} = 9$.
Step 3: Analysis
Substituting $D^{2} = 9$ into the denominator: $\frac{1}{9 + 1} \cosh 3x = \frac{1}{10} \cosh 3x$.
Step 4: Conclusion
Since the denominator is non-zero, this direct substitution yields the particular integral. Final Answer: (A)
For hyperbolic functions like $\cosh ax$, we substitute $D^{2} = a^{2}$ in the operator.
Step 2: Meaning
$P.I. = \frac{1}{D^{2} + 1} \cosh 3x$. Here $a = 3$, so $a^{2} = 9$.
Step 3: Analysis
Substituting $D^{2} = 9$ into the denominator: $\frac{1}{9 + 1} \cosh 3x = \frac{1}{10} \cosh 3x$.
Step 4: Conclusion
Since the denominator is non-zero, this direct substitution yields the particular integral. Final Answer: (A)
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