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Question

(A) Given equation is $xy = ae^x + be^{-x}$.First, differentiate with respect to $x$ using the product rule on the left side:$x \frac{dy}{dx} + y = ae

Vedclass · Accepted Answer

$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy = 0$

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(A) Given equation is $xy = ae^x + be^{-x}$.First, differentiate with respect to $x$ using the product rule on the left side:$x \frac{dy}{dx} + y = ae^x - be^{-x}$.Now, differentiate again with respect to $x$:$x \frac{d^2 y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = ae^x + be^{-x}$.Simplify the expression:$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = ae^x + be^{-x}$.Since $ae^x + be^{-x} = xy$, substitute this back into the equation:$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = xy$.Rearranging gives the final differential equation:$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy = 0$.

The differential equation obtained by eliminating the arbitrary constants $a$ and $b$ from $xy = ae^x + be^{-x}$ is

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