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

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.

• The equation in the title and the one in the body are different. The one in the body is the same as in your last post and still separable, the one in the title may not have a symbolic solution. Almost all ODE do not have a symbolic solution. Commented Mar 7 at 7:27
• Well, I am embarassed , Thank you for remind me about the equation, in the body the equation is right. Commented Mar 7 at 8:17

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.
• @IvánGM What do you think of my edit? I had trouble following it until I worked it out, which tells me the typical Diff. Eq. student, could very much use some elaboration. The $y^2$ is particularly something I don't think I've ever seen before. Commented Mar 7 at 10:13
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}