Question

Solve: $(3xy+y^2)dx+(x^2+xy)dy=0$, $y\left(1\right)=1$

• A. $x^{2}y(2x+y)=3$
• B. $y^{2}y(2x+y)=3$
• C. $y^{2}y(x+2y)=3$
• D. $x^{2}y(x+2y)=3$

Answer 1

The correct option is A $x^{2}y(2x+y)=3$

$(x^2+xy)\frac{dy}{dx}+y^2+3xy=0$
Substituting $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$(x^2+x^{2}v)(x\frac{dv}{dx}+v)+x^2{v}^2+3x^{2}v=0\Rightarrow x^2(x\frac{dv}{dx}+2v^2+(x\frac{dv}{dx}+4)v)=0\Rightarrow \frac{dv}{dx}=-\frac{2v^2+2v}{x(v+1)}=\frac{2(v+2)v}{x(v+1)}\Rightarrow \frac{\frac{dv}{dx}(v+1)}{v(v+2)}=-\frac{2}{x}$
Integrating both sides w.r.t x, we get
$\int \frac{\frac{dv}{dx}(v+1)}{v(v+2)}dx=\int -\frac{2}{x}dx\Rightarrow \frac{1}{2}\log (v+2)+\frac{1}{2}\log v=-2\log x+c\Rightarrow \frac{1}{2}\log (\frac{y}{x}+2)+\frac{1}{2}\log \frac{y}{x}=-2\log x+c$
Now $y\left(1\right)=1\Rightarrow x^{2}y(2x+y)=3$