Answer to Question #132209 in Differential Equations for Rouhish Ray

Question

Solve: (3ydx - 2xdy) + x^2y^-1(10ydx - 6xdy)=0

Answer 1

$(3ydx - 2xdy) + x^2y^{-1}(10ydx - 6xdy)=0$

$y(3ydx - 2xdy) + x^2(10ydx - 6xdy)=0$

$(3y^2+10x^2y)dx-(2xy+6x^3)dy=0$

$y(3y+10x^2)dx-2x(y+3x^2)dy=0$

$y(3y+10x^2)(2xdx)-4x^2(y+3x^2)dy=0$

$t=x^2, dt=2xdx$

$y(3y+10t)dt-4t(y+3t)dy=0$

$\frac{dy}{dt}=\frac{y(3y+10t)}{4t(y+3t)}$

This is an ODE of the homogeneous kind.

$y=tu, \frac{dy}{dt}=u+t\frac{du}{dt}$

Then:

$u+t\frac{du}{dt}=\frac{tu(3tu+10t)}{4t(tu+3t)}=\frac{u(3u+10)}{4(u+3)}$

$t\frac{du}{dt}=\frac{u(3u+10)}{4(u+3)}-u=\frac{3u^2+10u-4u^2-12}{4(u+3)}=\frac{-u^2+10u-12}{4(u+3)}$

$-\int\frac{4(u+3)}{u^2-10u+12}du=\int\frac{dt}{t}$

$-\frac{2}{13}((13+8\sqrt{13})ln(-u+\sqrt{13}+5)+(13-8\sqrt{13})ln(u+\sqrt{13}-5))=ln|u|+c$

$u=y/x^2$

Answer:

$-\frac{2}{13}((13+8\sqrt{13})ln(-y/x^2+\sqrt{13}+5)+(13-8\sqrt{13})ln(y/x^2+\sqrt{13}-5))=$

$=ln|y/x^2|+c$