Solve Differential Equation : $ (4xy^2) dx +(2x^2 +\frac{y}{x^2})dy= 0 $

Question

I am trying again and again to solve this but I got wrong answer and the lecturer told me is wrong, I don't know he's right or not. First of all , I have asked same question,someone told me to solve with separable variable , yes it is easy then , the answer is

$\sqrt[3]{e^{x^4}} - \sqrt[3]{c} = y $

And I don't know how to solve this because When I using exact method, I met the the complicated construction,

$$(4xy^2) dx + (2x^2y +\frac {y}{x^2})dy = 0 $$

$M = (4xy^2) dx$

$N = (2x^2y + \frac{y}{x^2})dy$

$\frac{dM}{dx} =8xy$

$\frac{dN}{dx} = 4xy-\frac{2y}{x^3} $

$R(x) = \frac{8xy - (4xy - \frac{2y}{x^3})} {2x^2y + \frac{y}{x^2}}$

$R(x) = \frac{4xy + \frac{2y}{x^3}}{2x^2y + \frac{y}{x^2}}$

If we see only R(x) is a good option but to integrated look like complicated so, the next step become

$ e^{\int\frac{4xy + \frac{2y}{x^3}}{2x^2y + \frac{y}{x^2}}} $

I have tried using someway to solve keep the answer doesn't meet except separable , maybe there are something I missed .

I would like appreciate if you help me to find the way and the answer, thanks.

Answer 1

I'm going to assume you want to solve the equation $$4xy^2 + \left(2x^2y +\frac {y}{x^2}\right)\frac{dy}{dx} = 0,$$ as in the body, by using an integrating factor. Let $M$ and $N$ be $M=4xy^2$, $N=2x^2y +\frac {y}{x^2}$. Then, after some calculations, $$ \dfrac{M_y-N_x}{N}=\dfrac{\dfrac{2y(2x^4+1)}{x^3}}{\dfrac{y(2x^4+1)}{x^2}}=\dfrac{2}{x}. $$ The fact that $\frac{2}{x}$ is a function only depending on $x$, is a sufficient condition to ensure the equation has an integrating factor only depending on $x$. Such a factor can be found as: $$ \mu (x) = \exp\left({\displaystyle\int \dfrac{2}{x}} dx\right) = x^2. $$ Multiplying the original equation by $\mu (x) = x^2$, it becomes an exact equation, whose solution is given by $$ y^2(2x^4+1)=c, $$ where $c$ is a (no negative) constant.

Answer 2

An alternative is separation of variables. You'll just have to multiply by $x^2/y^2$ on both sides: $$ \begin{aligned} 4xy^2\mathrm dx+\left(2x^2y+\frac{y}{x^2}\right)\mathrm dy&=0\\ 4x^3\mathrm dx+\left(2x^4+1\right)\frac{\mathrm dy}{y}&=0\\ \frac{1}{2}\int\frac{8x^3}{1+2x^4}\mathrm dx+\int\frac{\mathrm dy}{y}&=0\\ \frac{1}{2}\ln(1+2x^4)+\ln y&=C'\\ \implies y&=\frac{C}{\sqrt{1+2x^4}} \end{aligned}$$