how do i solve this problem here? someone please help me
(2y^2+5x)dx+3xydy=0 solve
This equation can be made exact by finding an integrating factor
Let $\displaystyle \phi(x)$ be a differentable function of x
multiplying the equation by phi we get
$\displaystyle (2y^2+5x)\phi(x) dx+3xy \phi(x) dy=0 $
Remember to be exact that if we have an eqaution of the form
$\displaystyle Mdx+Ndy=0$ it is exact if
$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
$\displaystyle 4y\phi(x)=3x\phi(x)+3xy\frac{d\phi}{dx} \iff y\phi(x)=3xy\frac{d\phi}{dx} \iff \frac{1}{x}\cdot \frac{1}{x}dx=\frac{1}{\phi}d\phi$
Now solving for phi we get
$\displaystyle \frac{1}{3}\ln(x)=\ln(\phi) \iff \phi(x)=x^{1/3}$
So using this we get
$\displaystyle (2x^{1/3}y^2+5x^{4/3})dx+3x^{4/3}ydy=0$
This is now exact and can be solve using that method.
I hope this helps.