Find the particular solution of the differential equation $x^2 d y=\left(2 x y+y^2\right) d x$ given that $y=1$ when $x=1$

The given differential equation is $x^{2} dy = (2xy + y^{2}) dx$, which can be written as $\dfrac{dy}{dx} = \dfrac{2xy + y^{2}}{x^{2}}$.

This is a homogeneous differential equation. Let $y = vx$, then $\dfrac{dy}{dx} = v + x \dfrac{dv}{dx}$.

Substituting these into the equation: $v + x \dfrac{dv}{dx} = \dfrac{2x(vx) + (vx)^{2}}{x^{2}} = \dfrac{2vx^{2} + v^{2}x^{2}}{x^{2}} = 2v + v^{2}$.

Rearranging the terms: $x \dfrac{dv}{dx} = 2v + v^{2} - v = v^{2} + v$.

Separating the variables: $\dfrac{dv}{v^{2} + v} = \dfrac{dx}{x}$, which is $\dfrac{dv}{v(v+1)} = \dfrac{dx}{x}$.

Using partial fractions: $\left( \dfrac{1}{v} - \dfrac{1}{v+1} \right) dv = \dfrac{dx}{x}$.

Integrating both sides: $\ln|v| - \ln|v+1| = \ln|x| + C$, which simplifies to $\ln \left| \dfrac{v}{v+1} \right| = \ln|x| + C$.

Exponentiating: $\dfrac{v}{v+1} = Cx$.

Substituting $v = \dfrac{y}{x}$: $\dfrac{y/x}{y/x + 1} = Cx \Rightarrow \dfrac{y}{y+x} = Cx \Rightarrow y = Cx(y+x)$.

Given $y=1$ when $x=1$: $1 = C(1)(1+1) \Rightarrow 1 = 2C \Rightarrow C = \dfrac{1}{2}$.

Substituting $C = \dfrac{1}{2}$ into the equation: $y = \dfrac{1}{2}x(y+x) \Rightarrow 2y = x(x+y)$.

Answer: $2 y=x(x+y)$