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How would you solve the differential equation; $$~(4x^2+4y^2) dy + xy dx = 8~?$$ Can you even do this question by substituting $~V= \dfrac{y}{x}~$ and solving from there?

Trying the get $~\dfrac{dy}{dx}~$ by itself is a tough ask too, so is this question insanely difficult? Or am I missing something simple?

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    $\begingroup$ This equation is impossible. On the left side you have an infinitesimal or a linear functional, on the right you have a constant. These are different types of objects that never can be equal. $\endgroup$ Sep 19, 2019 at 7:46
  • $\begingroup$ I got this on a test... So for me to solve it, the right hand side would need to equal 0? $\endgroup$ Sep 19, 2019 at 7:59
  • $\begingroup$ Yes. Could the 8 be a misprint or other deformation of a $0$ on the paper? $\endgroup$ Sep 19, 2019 at 8:06
  • $\begingroup$ Hopefully, that was the only question that I couldn't answer! $\endgroup$ Sep 19, 2019 at 8:45

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It hasn't any solution since differentials equals to 8!

$dx,dy$ is near $0$ for every $x,y$, so the equality can't hold!

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If the right side were zero, you could group the left side by degrees $$ [4x^2\,dy+xy\,dx] + 4y^2\,dy=0 $$ and contract to complete differentials $$ 0=\frac12y^{-7}\,d(x^2y^8)+4y^2\,dy=\frac12y^{-7}d\left(x^2y^8+\frac45y^{10}\right) $$ This means that $10y^7$ is an integrating factor and $$ 5x^2y^8+4y^{10}=C $$ are the solution curves.

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