Question

Solve: $(2x+3y)dx+xdy=0;$ when $x=1,y=0$.

Answer 1

$(2x+3y)dx+xdy=0\Rightarrow \frac{dy}{dx}=-\frac{-(2x+3y)}{x}⟶\left(1\right)$

since the equation is homogeneous

Let, $y=vx\Rightarrow \frac{dy}{dx}=x\frac{dv}{dx}+v$

putting the above value in equation (1)

$\Rightarrow x\frac{dv}{dx}+v=\frac{-(2x+3vx)}{x}\Rightarrow x\frac{dv}{dx}=-(2+4v)\Rightarrow \frac{dv}{(2+4v)}=-\frac{dx}{x}\Rightarrow \int \frac{dv}{(2+4v)}=-\int \frac{dx}{x}\Rightarrow \frac{1}{4}log(2+4v)=-\log x+\log C\Rightarrow \log {(2+4v)}^{\frac{1}{4}}+\log x=\log C\Rightarrow \log x{(2+4v)}^{\frac{1}{4}}=\log C\Rightarrow x{(2+4v)}^{\frac{1}{4}}=C\Rightarrow x{(2+\frac{4y}{x})}^{\frac{1}{4}}=C$