Answer to Question #209121 in Differential Equations for haha

Question

Given the equation below, solve for the general solutions.

w = dependent variable, v = independent variable

1. w (4v + w) dv - 2 (v^2 - w) dw = 0

Answer 1

$M_w=4v+2w\neq N_v=-4v$

so the equation is not exact.

$w(v)=\displaystyle{\sum_{n=0}^{\infin}}a_nv^n$

$w'(v)=\displaystyle{\sum_{n=1}^{\infin}}na_nv^{n-1}$

$4v\displaystyle{\sum_{n=0}^{\infin}}a_nv^n+(\displaystyle{\sum_{n=0}^{\infin}}a_nv^n)^2-2v^2+2\displaystyle{\sum_{n=0}^{\infin}}a_nv^n\cdot \displaystyle{\sum_{n=1}^{\infin}}na_nv^{n-1}=0$

$4a_1v^2+(2a_0a_2+a_1^2)v^2+(6a_0a_3+4a_1a_2+2a_2a_1)v^2=2v^2$

$4a_1+2a_0a_2+a_1^2+6a_0a_3+6a_1a_2=2$

$a_0=0$

Then:

$a_2=\frac{2-4a_1-a_1^2}{6a_1}$

$y(x)=a_1x+\frac{2-4a_1-a_1^2}{6a_1}x^2$