Answer to Question #179739 in Differential Equations for kyle panoringan

Question

(x2 + y2)dx + x(x − 2y)dy = 0

Answer 1

(x²+ y²)dx + x(x − 2y)dy = 0

=> $\frac{dy}{dx}=\frac{x^2+y^2}{2xy-x²}$

This is a homogeneous differential equation

Let us put y = vx and so $\frac{dy}{dx}=v+ x\frac{dv}{dx}$

The equation transforms to

$v+ x\frac{dv}{dx}$ = $\frac{x^2+v^2x^2}{2vx^2-x²}$

=> $v+ x\frac{dv}{dx}$ = $\frac{1+v^2}{2v-1}$

=> $x\frac{dv}{dx}$ = $\frac{1+v^2}{2v-1} - v$

=> $x\frac{dv}{dx}$ = $\frac{1+v-v^2}{2v-1}$

=> $\frac{2v-1} {1+v-v^2} dv$ = $\frac{dx} {x}$

=> $\frac{2v-1} {v^2-v-1} dv$ = $-\frac{dx} {x}$

=> $\int\frac{2v-1} {v^2-v-1} dv =$ $-\int\frac{dx} {x}$ +ln|C|

=> ln|v²-v-1} = - ln|x| + ln|C| [since $\int\frac{f'(x)}{f(x)} dx = ln f(x)]$

=> ln|v²-v-1| + ln|x| = ln|C|

=> ln|v²x-vx-x} = ln|C|

=> $ln| \frac{y²}{x} -y -x| = ln|C|$

=> $ln| \frac{y²-xy -x²}{x}| = ln|C|$

=> $| \frac{y²-xy -x²}{x}| = |C|$

=> $(y²-xy -x²)²= C²x²$