$ (x^2+xy^4)dx+ 2y^3dy=0 $

Question

Prove that $ {e^x}^2 $ is an integrating factor of the equation

$$ (x^2+xy^4)dx+ 2y^3dy=0 $$

and hence solve it.

I have tried to solve the equation, however, I am getting the $ e^{2x^2} $ as the integrating factor. What have I done wrong?

Answer 1

Since $N=2y^3$, we have

$$\frac{\large{\frac{\partial M}{\partial y}}-\frac{\partial N}{\partial x}}{N}=\frac{4xy^3-0}{2y^3}=2x$$

therefore the integrating factor is

$$\text{IF}=\large{e^{\int f(x)\,dx}}=\large{e^{\int 2x\,dx}}=\large{e^{x^2}}$$

Answer 2

$$(x^2+xy^4)dx+ 2y^3dy=0$$ $$x^2dx+(xy^4dx+ 2y^3dy)=0$$ $$x^2dx+\frac 12(2xy^4dx+ 4y^3dy)=0$$ $$x^2dx+\frac 12(y^4dx^2+ dy^4)=0$$ Integrating factor $\mu =e^{x^2}$ $$x^2e^{x^2}dx+\frac 12(e^{x^2}y^4dx^2+ e^{x^2}dy^4)=0$$ $$x^2e^{x^2}dx+\frac 12(de^{x^2}y^4)=0$$ Integrate: $$\int x^2e^{x^2}dx+\frac 12e^{x^2}y^4=C$$ You have an error function for the integral.