Mathematics - Differential Equation Question with Solution | TestHub

Given $\frac{dy}{dx}=\frac{x{y}^2+y}{x}$

$\frac{xdy-ydx}{y^2}=xdx$

$-d\left(\frac{x}{y}\right)=xdx$

$-\frac{x}{y}=\frac{x^2}{2}+c$

$\because$ Curve intersects the line $x+2y=4$ at $x=-2\Rightarrow$ point of intersection is $\left(-2,3\right)$. 

$\therefore$ Curve passes through $\left(-2,3\right)$. 

$\Rightarrow \frac{2}{3}=2+c\Rightarrow c=-\frac{4}{3}$

$\Rightarrow \frac{-x}{y}=\frac{x^2}{2}-\frac{4}{3}$

Now put $\left(3,y\right)$ we get, 

$\Rightarrow \frac{-3}{y}=\frac{19}{6}$

$\Rightarrow y=\frac{-18}{19}$