If x frac{dy}{dx} + y = frac{x f(xy)}{f'(xy)} | vedclass

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(A) Given the differential equation: $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.Substituting

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$k e^{x^2 / 2}$

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(A) Given the differential equation: $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.Substituting this into the equation,we get: $\frac{d(xy)}{dx} = \frac{x f(xy)}{f'(xy)}$.Rearranging the terms to separate the variables $xy$ and $x$: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.This yields: $\ln |f(xy)| = \frac{x^2}{2} + C$.Taking the exponential of both sides: $|f(xy)| = e^{\frac{x^2}{2} + C} = e^C \cdot e^{x^2 / 2}$.Letting $k = e^C$,we get: $|f(xy)| = k e^{x^2 / 2}$.

If $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$,then $|f(xy)|$ is equal to

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