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

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

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

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(A) Given the differential equation: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.We know that $x dy + y dx = d(xy)$.Rearranging the given equation: $x dy + y dx = x \frac{f(xy)}{f'(xy)} dx$.This is not quite right,let's rewrite: $\frac{dy}{dx} + \frac{y}{x} = \frac{f(xy)}{f'(xy)}$.Multiply by $x$: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.Note that $x \frac{dy}{dx} + y = \frac{d}{dx}(xy)$.So,$\frac{d(xy)}{dx} = x \frac{f(xy)}{f'(xy)}$.Rearranging the terms: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.$\ln|f(xy)| = \frac{x^2}{2} + C$.Taking the exponential of both sides: $f(xy) = e^{\frac{x^2}{2} + C} = k e^{x^2/2}$,where $k = e^C$.

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

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