# D Y D X + Y X = Y 2 X 2 - Mathematics

Sum

$\frac{dy}{dx} + \frac{y}{x} = \frac{y^2}{x^2}$

#### Solution

We have,

$\frac{dy}{dx} + \frac{y}{x} = \frac{y^2}{x^2}$

$\Rightarrow \frac{dy}{dx} = \left( \frac{y}{x} \right)^2 - \frac{y}{x}$

Putting y = vx, we get

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

$\therefore v + x\frac{dv}{dx} = v^2 - v$

$\Rightarrow x\frac{dv}{dx} = v^2 - 2v$

$\Rightarrow \frac{1}{v^2 - 2v} dv = \frac{1}{x}dx$

Integrating both sides, we get

$\int\frac{1}{v^2 - 2v} dv = \int\frac{1}{x}dx$

$\Rightarrow \int\frac{1}{v^2 - 2v + 1 - 1} dv = \int\frac{1}{x}dx$

$\Rightarrow \int\frac{1}{\left( v - 1 \right)^2 - \left( 1 \right)^2} dv = \int\frac{1}{x}dx$

$\Rightarrow \frac{1}{2}\log \left| \frac{v - 1 - 1}{v - 1 + 1} \right| = \log x + \log C$

$\Rightarrow \log \left| \left( \frac{v - 2}{v} \right)^\frac{1}{2} \right| = \log Cx$

$\Rightarrow \log \left| \left( \frac{\frac{y}{x} - 2}{\frac{y}{x}} \right)^\frac{1}{2} \right| = \log Cx$

$\Rightarrow \log \left| \left( \frac{y - 2x}{y} \right)^\frac{1}{2} \right| = \log Cx$

$\Rightarrow \left( \frac{y - 2x}{y} \right)^\frac{1}{2} = Cx$

$\Rightarrow \frac{y - 2x}{y} = C^2 x^2$

$\Rightarrow y - 2x = k x^2 y,\text{ where }k = C^2$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 37 | Page 146