Solve the differential equation (if exact): $(5x + 4y)dx + (4x - 8y^3)dy = 0$ I need to test this to see if it is exact. My conclusion is the it is indeed exact because $My = 4 = Nx$. then I integrated $Mdx$ and $Ndy$.
$$
\int Mdx = \int (5x+4y)\, dx = (5/2)x^2 + 4xy +f(y)
$$
$$\int Ndy = \int (4x-8y^3)\, dy = 4xy-2y^4 + g(x) 
$$
What am I supposed to conclude in regards to the $4xy$ being present in both equations? Does this mean that my final answer is $$(5/2)x^2 - 2y^4 + 4xy + C = 0$$ or $$(5/2)x^2 - 2y^4 + 8xy + C = 0$$
??
 A: Here is a rule which almost always works: (You can differentiate the solution to check). Integrate $M$ with respect to $x$ keeping $y$ as a constant and integrate only those terms of $N$ which do not involve $x$ with respect to $y$. Sum the result and equate to a constant.
So as $\displaystyle\int_{y\mbox{ constant}}Mdx=\int(5x+4y)dx=\frac{5x^2}{2}+4xy$, and $\displaystyle\int\mbox{(Terms of $N$ free of $x$) } dy=-8\int y^3dy=-2y^4$ we have $\frac{5x^2}{2}+4xy-2y^4=c$ as the solution.
A: In this case, you may solve it directly:
\begin{align}
(5x+4y)\,dx + (4x-8y^3)\,dy&=5x\,dx + (4y\,dx+4x\,dy) - 8y^3\,dy=\\
&=d(5x^2/2) + d(4xy)-d(2y^4),
\end{align}
and
$$5x^2/2 + 4xy-2y^4=C.$$
Or we can take
$$
\int Mdx = \int (5x+4y)\, dx = (5/2)x^2 + 4xy +f(y)
$$
and differentiate it:
$$
\frac{d}{dy}\int Mdx = 4x +f'(y) = 4x-8y^3 \Longrightarrow f'(y) = -8y^3
$$
and then $f(y) = -2y^4 + C$.
A: $\bf hint: $ differentiate the right hand side of $(1)$ with respect to $y$ and set it to $4x - 8y^3.$   that should allow you to find $f$ unto a constant.
A: You could differentiate your answer, substitute back into your equation and find out yourself which one it is.
