Solve the following differential equation:
$x^{2} d y + (x y + y^{2}) d x = 0$ , when $x = 1$ and $y = 1$

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Solution

The given differential equation is
$x^{2} d y + (x y + y^{2}) d x = 0$
$\Rightarrow \frac{d y}{d x} = - \frac{y (x + y)}{x^{2}}$
which is homogeneous as each of the functions in the numerator and the denominator is a homogeneous function of degree $2$ .
Putting $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , the given equation becomes
$v + x \frac{d v}{d x} = \frac{- v x (x + v x)}{x^{2}}$
$x \frac{d v}{d x} = - v (1 + v) - v = - (2 v + v^{2}) = - v (v + 2)$
$\Rightarrow \frac{d v}{v (v + 2)} = - \frac{1}{x} d x$
$∴ \int \frac{d v}{v (v + 2)} = - \int \frac{d x}{x}$
$\int \frac{1}{2} (\frac{1}{v} - \frac{1}{v + 2}) d v = - \int \frac{d x}{x}$
$\Rightarrow \frac{1}{2} log | v | - \frac{1}{2} log | v + 2 | = - log | x | + log c$
$\Rightarrow log ∣ ∣ ∣ \frac{x \sqrt{v}}{\sqrt{v + 2}} ∣ ∣ ∣ = log c$
$\Rightarrow \frac{x \sqrt{v}}{\sqrt{v + 2}} = c \Rightarrow \frac{x \sqrt{y}}{\sqrt{y + 2}} = c$ (putting $v = \frac{y}{x}$ )
$\Rightarrow$ $x^{2} y = A (y + 2 x) . . . . . . (i)$
Where $A = c^{2}$
when $x = 1$ and $y = 1$ , equation $(i)$ becomes
$1 = A (1 + 2) \Rightarrow$ $A = \frac{1}{3}$
Putting the value of $A$ in equation $(i)$ , we get
$3 x^{2} y = (y + 2 x)$

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