Answer:
[tex]\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C[/tex]
Step-by-step explanation:
Given differential equation,
[tex](x^2 + y^2) dx + (x^2 - xy) dy = 0[/tex]
[tex]\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)[/tex]
Let y = vx
Differentiating with respect to x,
[tex]\frac{dy}{dx}=v+x\frac{dv}{dx}[/tex]
From equation (1),
[tex]v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}[/tex]
[tex]v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}[/tex]
[tex]v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}[/tex]
[tex]v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}[/tex]
[tex]x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v[/tex]
[tex]x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}[/tex]
[tex]x\frac{dv}{dx}=\frac{v+1}{v-1}[/tex]
[tex]\frac{v-1}{v+1}dv=\frac{1}{x}dx[/tex]
Integrating both sides,
[tex]\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx[/tex]
[tex]\int{\frac{v-1+1-1}{v+1}}dv=lnx + C[/tex]
[tex]\int{1-\frac{2}{v+1}}dv=lnx + C[/tex]
[tex]v-2ln(v+1)=lnx+C[/tex]
Now, y = vx ⇒ v = y/x
[tex]\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C[/tex]
