Find the particular solution of the differential equation satisfy… - Vidyadip

Question

Find the particular solution of the differential equation satisfying the given conditions:
$x^2dy + (xy + y^2) dx = 0 ; y = 1$ when $x = 1.$

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Answer

The given differential equation can be written as
$\frac{\text{dy}}{\text{dx}}+\frac{\text{xy+y}^{2}}{\text{x}^{2}}=0\text{ }\cdot\text{ }\text{Let y = vx}\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\text{v + x}\frac{\text{dv}}{\text{dx}}+\text{(v + v}^{2})=0$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=-\text{ v (2 + v)}$
$\text{OR }\frac{\text{dv}}{\text{v(2+v)}}=-\frac{\text{dx}}{\text{x}}$
$\text{OR }\int\Bigg(\frac{1}{\text{v}}-\frac{1}{\text{2 + v}}\Bigg)\text{dx}=-2\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\log\frac{\text{v}}{\text{v+2}}=\log\frac{\text{c}}{\text{x}^{2}}$
$\text{OR }\frac{\text{y}}{\text{y+2}}=\frac{\text{c}}{\text{x}^{2}}$
when $x = 1, y = 1 \Rightarrow\text{c}=\frac{1}{3}$
$\therefore$ The solution becomes
$y + 2x = 3x^2y.$

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