Question

Solve the equation:
$(x^2+3xy+y^2)dx-x^{2}dy=0$, given that $y=0$ and $x=1$.

Answer 1

$(x^2+3xy+y^2)dx-x^{2}dy=0$

$\frac{(x^2+3xy+y^2)}{x^2}=\frac{dy}{dx}$

$1+3\left(\frac{y}{x}\right)+(\frac{y}{x}{)}^2=\frac{dy}{dx}$

let $\frac{y}{x}=t----\left(1\right)$

y = xt

diff. w.r.t x

$\frac{dy}{dx}=t+x\frac{dt}{dx}$

$\left(i\right)\Rightarrow 1+3\left(t\right)+(t{)}^2=t+x\frac{dt}{dx}$

$t^2+2t+1=x\frac{dt}{dx}$

$\frac{dx}{x}=\frac{dt}{(t+1{)}^2}$

integrating both sides

$\int \frac{dx}{x}=\int \frac{dt}{(t+1{)}^2}$

$ln\left|x\right|=\frac{-1}{(t+1)}+c$

$ln\left|x\right|=\frac{-1}{(\frac{y}{x}+1)+c}$

$ln\left|x\right|=\frac{-x}{x+y}+c$

at x = 1, y = 0

$lnl=\frac{-1}{1+0}+c$

$0=-1+c\Rightarrow c=1$

$ln\left|x\right|=\frac{-x}{x+y}+1$