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Justify that a differential equation of the form: $$[y+xf(x^2+y^2)]dx+[yf(x^2+y^2)-x]dy=0$$ , where $f(x^2+y^2)$ is an arbitrary function of $(x^2+y^2)$, is not an exact differential equation and $1/(x^2+y^2)$ is an integrating factor for it. Hence, solve this differential equation for $$f(x^2+y^2)=(x^2+y^2)^2$$

I am unable to solve after making it an exact equation, having difficulty in the integration of this question.

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    $\begingroup$ Please refrain from posting questions as pictures: use MathJax to format questions, and also add your working. $\endgroup$ Commented Oct 29, 2020 at 13:31

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$$(y+xf(x^2+y^2))dx+(yf(x^2+y^2)-x)dy=0$$ $$f(x^2+y^2))(xdx+ydy)+ydx-xdy=0$$ $$\dfrac 12f(x^2+y^2))(d(x^2+y^2))+ydx-xdy=0$$ Divide by $x^2+y^2$: $$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}+\dfrac {ydx-xdy}{x^2+y^2}=0$$ Note that: $$d \arctan (y/x)=\dfrac {xdy-ydx}{x^2+y^2}$$ $$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}-d(\arctan (y/x)) =0$$ Integrate for the given function $f$.

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