Find the particular solution of the differential equation satisfying the given conditions: x² dy + (xy + y² )dx = 0; y = 1 when x = 1.

$$\begin{gathered} {\mathrm{x}}^2 \mathrm{dy} + \left( \mathrm{xy} + {\mathrm{y}}^2 \right) \mathrm{dx} = 0 \\ {\mathrm{x}}^2 \mathrm{dy} =- \left( \mathrm{xy} + {\mathrm{y}}^2 \right) \mathrm{dx} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{- \left( \mathrm{xy} + {\mathrm{y}}^2 \right)}{{\mathrm{x}}^2} ..........\left(\mathrm{i}\right) \end{gathered}$$

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, 

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}} \left( \mathrm{y} \right) = \frac{{\displaystyle \mathrm{d}}}{{\displaystyle \mathrm{dx}}} \left( \mathrm{vx} \right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{v} + \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} \\ \mathrm{substituting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{y} \mathrm{and} \frac{{\displaystyle \mathrm{dy}}}{{\displaystyle \mathrm{dx}}} \mathrm{in} \mathrm{equation} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get}: \\ \mathrm{v} + \mathrm{x} \frac{{\displaystyle \mathrm{dv}}}{{\displaystyle \mathrm{dx}}} = \frac{- \left[ \mathrm{x} \times \mathrm{vx} + {\left( \mathrm{vx} \right)}^2 \right]}{{\mathrm{x}}^2} \\ \Rightarrow \mathrm{x} \frac{{\displaystyle \mathrm{dv}}}{{\displaystyle \mathrm{dx}}} = - {\mathrm{v}}^2 - 2\mathrm{v} = -\mathrm{v} \left( \mathrm{v} + 2 \right) \\ \Rightarrow \frac{\mathrm{dv}}{\mathrm{v} \left( \mathrm{v} + 2 \right)} = - \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow \frac{1}{2} \left[ \frac{1}{\mathrm{v}} - \frac{1}{\mathrm{v} + 2} \right] \mathrm{dv} = - \frac{\mathrm{dx}}{\mathrm{x}} \end{gathered}$$

Integrating both sides, we get:

$$\begin{gathered} \frac{1}{2} \left[ \log \mathrm{v} - \log \left( \mathrm{v} + 2 \right)\right] = - \log \mathrm{x} + \log \mathrm{C} \\ \Rightarrow \frac{1}{2} \log \left( \frac{\mathrm{v}}{\mathrm{v} + 2 }\right) = \log \frac{\mathrm{C}}{\mathrm{x}} \\ \Rightarrow \frac{{\displaystyle \mathrm{v}}}{{\displaystyle \mathrm{v} + 2 }} = {\left(\frac{{\displaystyle \mathrm{C}}}{{\displaystyle \mathrm{x}}}\right)}^2 \\ \mathrm{Substituting} \mathrm{v} = \frac{\mathrm{y}}{\mathrm{x}} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{\mathrm{y}}{\mathrm{x}}}}{{\displaystyle \frac{\mathrm{y}}{\mathrm{x}}} + 2} = {\left(\frac{\mathrm{C}}{\mathrm{x}}\right)}^2 \\ \Rightarrow \frac{\mathrm{y}}{\mathrm{y} + 2\mathrm{x}} = \frac{{\mathrm{C}}^2}{{\mathrm{x}}^2} \\ \Rightarrow \frac{{\mathrm{x}}^2\mathrm{y}}{\mathrm{y} + 2\mathrm{x}} = \mathrm{D} ............\left(\mathrm{ii}\right) \end{gathered}$$

Now, it is given that  y = 1  at  x = 1.

$$\begin{gathered} \Rightarrow \frac{1}{1 + 2} = \mathrm{D} \Rightarrow \mathrm{D} = \frac{1}{3} \\ \mathrm{Substituting} \mathrm{D} = \frac{1}{3} \mathrm{in} \mathrm{equation} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{x}}^2\mathrm{y}}{\mathrm{y} + 2\mathrm{x}} = \frac{1}{3} \Rightarrow \mathrm{y} + 2\mathrm{x} = 3{\mathrm{x}}^2\mathrm{y} \\ \mathrm{So}, \mathrm{the} \mathrm{required} \mathrm{solution} \mathrm{is} \mathrm{y} + 2\mathrm{x} = 3{\mathrm{x}}^2\mathrm{y} \end{gathered}$$