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  • The differential equation for y = A cos alpha x + B sin alpha x are arbitrary constants is

The differential equation for \mathrm{y}=\mathrm{A} \cos \alpha \mathrm{x}+\mathrm{B} \sin \alpha \mathrm{x}, where \mathrm{A} and \mathrm{B}
are arbitrary constants is
\\A. \frac{d^{2} y}{d x^{2}}-\alpha^{2} y=0 \\\\\mathrm{B} \frac{d^{2} y}{d x^{2}}+\alpha^{2} y=0 \\\\C. \frac{d^{2} y}{d x^{2}}+\alpha y=0 \\\\D. \frac{d^{2} y}{d x^{2}}-\alpha y=0

Answers (1)

Let us find the differential equation by differentiating y with respect to x twice
Twice because we have to eliminate two constants \mathrm{A}$ and $\mathrm{B}$ \\.

\mathrm{y}=\mathrm{A} \cos \alpha \mathrm{x}+\mathrm{B} \sin \alpha \mathrm{x}
Differentiating,
$$ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{A} \alpha \sin \alpha \mathrm{x}+\mathrm{B} \alpha \cos \alpha \mathrm{x} $$
Differentiating again
$$ \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-A \alpha^{2} \cos \alpha x-B \alpha^{2} \sin \alpha x \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\alpha^{2}(A \cos \alpha x+B \sin \alpha x) $$

\\ \text { But } y=A \cos a x+B \sin a x \\ \quad \Rightarrow \frac{d^{2} y}{d x^{2}}=-\alpha^{2} y \\ \Rightarrow \frac{d^{2} y}{d x^{2}}+\alpha^{2} y=0

Option B is correct.

Posted by

infoexpert22

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