# How would you solve the differential equation; $~(4x^2+4y^2) dy + xy dx = 8~$?

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?

• 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. Sep 19, 2019 at 7:46
• I got this on a test... So for me to solve it, the right hand side would need to equal 0? Sep 19, 2019 at 7:59
• Yes. Could the 8 be a misprint or other deformation of a $0$ on the paper? Sep 19, 2019 at 8:06
• Hopefully, that was the only question that I couldn't answer! Sep 19, 2019 at 8:45

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