$x^2$ dy + (xy + $y^2$) dx = 0 $x^2$ dy = - (xy + $y^2$) dx ⇒ $\frac{dy}{dx}$ = - $\frac{xy+y^2}{x^2}$ ... (1) This is a homogeneous differential equation. Such type of equations can be reduced to variable separable form by the substitution y = vx. Differentiating w.r.t. x we get, $\frac{d}{dx}$ (y) = $\frac{d}{dx}$ (vx) ⇒ $\frac{dy}{dx}$ = v + x $\frac{dv}{dx}$ Substituting the value of y and $\frac{dy}{dx}$ in equation (1), we get : v + x $\frac{dv}{dx}$ = - $\frac{x \times vx + (vx)^2}{x^2}$ ⇒ x $\frac{dv}{dx}$ = - $v^2$ - 2v = - v (v + 2) ⇒ $\frac{dv}{v(v+2)}$ = - $\frac{dx}{x}$ ⇒ $\frac{1}{2}\left[\frac{1}{x}-\frac{1}{v+2}\right]$ dv = - $\frac{dx}{x}$ Integrating both sides, we get: $\frac{1}{2}$ [log v log(v + 2)] = - log x + logC ⇒ $\frac{1}{2}\log\left(\frac{v}{v+2}\right)$ = log $\frac{C}{x}$ ⇒ $\frac{v}{v+2}$ = $\left(\frac{C}{x}\right)^2$ Substituting v = $\frac{y}{x}$ ⇒ $\frac{\frac{y}{x}}{\frac{y}{x}+2}$ = $\left(\frac{C}{x}\right)^2$ ⇒ $\frac{y}{y+2x}$ = $\frac{C^2}{x^2}$ ⇒ $\frac{x^2y}{y+2x}$ = D ... (2) Now, it is given that y = 1 at x = 1. ⇒ $\frac{1}{1+2}$ = D ⇒ D = $\frac{1}{3}$ Substituting D = $\frac{1}{3}$ in equation (2), we get $\frac{x^2y}{y+2x}$ = $\frac{1}{3}$ ⇒ y + 2x = $3x^2y$ So, the required solution is y + 2x = $3x^2y$