Answer to Question #292676 in Differential Equations for Nice

Question

Solve the Non-exact Equation

v(u2+v2)du - u(u2+2v2)dv

Answer 1

Solution:

Given, $v(u^2+v^2)du - u(u^2+2v^2)dv=0$

Replace u with x and v with y.

$y(x^2+y^2)dx - x(x^2+2y^2)dy=0 \\ \Rightarrow \dfrac{dy}{dx}=\dfrac{y(x^2+y^2)}{x(x^2+2y^2)}\ ...(i)$

Put $y=vx\ ...(ii)$

$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\ ...(iii)$

From (i), (ii) and (iii).

$v+ \dfrac{dv}{dx}=\dfrac{vx(x^2+v^2x^2)}{x(x^2+2v^2x^2)} \\ \Rightarrow \dfrac{dv}{dx}=\dfrac{v(1+v^2)}{(1+2v^2)}-v \\ \Rightarrow \dfrac{dv}{dx}=\dfrac{v+v^3-v-2v^3}{(1+2v^2)} \\ \Rightarrow \dfrac{dv}{dx}=\dfrac{-v^3}{(1+2v^2)} \\ \Rightarrow \dfrac{(1+2v^2)dv}{v^3}=-dx \\ \Rightarrow (\dfrac{1}{v^3}+\dfrac{2}{v})dv=-dx$

On integrating both sides,

$-\dfrac{1}{2v^2}+2\log v=-x+C \\ \Rightarrow-\dfrac{1}{2(\frac yx)^2}+2\log \frac yx=-x+C \\ \Rightarrow-\dfrac{x^2}{2y^2}+2\log \dfrac yx=-x+C$

Now, replace u with x and v with y back

$-\dfrac{u^2}{2v^2}+2\log \dfrac vu=-u+C$