${\mathit{x}}^2$ dy + (xy + ${\mathit{y}}^2$) dx = 0
${\mathit{x}}^2$ dy = - (xy + ${\mathit{y}}^2$) dx
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = - $\frac{\left(\mathit{x}\mathit{y}+{\mathit{y}}^2\right)}{{\mathit{x}}^2}$ ... (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,
$\frac{\mathit{d}}{\mathit{d}\mathit{x}}$ (y) = $\frac{\mathit{d}}{\mathit{d}\mathit{x}}$ (vx) ⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = v + x $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{x}}$
Substituting the value of y and $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ in equation (1), we get :
v + x $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{x}}$ = - $\frac{\left[\mathit{x}\times \mathit{v}\mathit{x}+{\left(\mathit{v}\mathit{x}\right)}^2\right]}{{\mathit{x}}^2}$
⇒ x $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{x}}$ = - ${\mathit{v}}^2$ - 2v = - v (v + 2)
⇒ $\frac{\mathit{d}\mathit{v}}{\mathit{v}\left(\mathit{v}+2\right)}$ = - $\frac{\mathit{d}\mathit{x}}{\mathit{x}}$
⇒ $\frac{1}{2}\left[\frac{1}{\mathit{x}}-\frac{1}{\mathit{v}+2}\right]$ dv = - $\frac{\mathit{d}\mathit{x}}{\mathit{x}}$
Integrating both sides, we get:
$\frac{1}{2}$ [log v log(v + 2)] = - log x + logC
⇒ $\frac{1}{2}\log \left(\frac{\mathit{v}}{\mathit{v}+2}\right)$ = log $\frac{\mathit{C}}{\mathit{x}}$
⇒ $\frac{\mathit{v}}{\mathit{v}+2}$ = ${\left(\frac{\mathit{C}}{\mathit{x}}\right)}^2$
Substituting v = $\frac{\mathit{y}}{\mathit{x}}$
⇒ $\frac{\frac{\mathit{y}}{\mathit{x}}}{\frac{\mathit{y}}{\mathit{x}}+2}$ = ${\left(\frac{\mathit{C}}{\mathit{x}}\right)}^2$
⇒ $\frac{\mathit{y}}{\mathit{y}+2\mathit{x}}$ = $\frac{{\mathit{C}}^2}{{\mathit{x}}^2}$
⇒ $\frac{{\mathit{x}}^2\mathit{y}}{\mathit{y}+2\mathit{x}}$ = 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
$\frac{{\mathit{x}}^2\mathit{y}}{\mathit{y}+2\mathit{x}}$ = $\frac{1}{3}$ ⇒ y + 2x = $3{\mathit{x}}^2\mathit{y}$
So, the required solution is y + 2x = $3{\mathit{x}}^2\mathit{y}$