Question

x² dy + y (x + y) dx = 0

Answer 1

We have, 
\[ x^2 dy + y\left( x + y \right) dx = 0\]
\[ \Rightarrow x^2 dy = - y\left( x + y \right) dx\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- y\left( x + y \right)}{x^2}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{- vx\left( x + vx \right)}{x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = - v\left( 1 + v \right)\]
\[ \Rightarrow x\frac{dv}{dx} = - v - v - v^2 \]
\[ \Rightarrow x\frac{dv}{dx} = - \left( v^2 + 2v \right)\]
\[ \Rightarrow \frac{dv}{\left( v^2 + 2v \right)} = - \frac{dx}{x}\]
\[ \Rightarrow \frac{dv}{v\left( v + 2 \right)} = - \frac{dx}{x}\]
Integrating both sides, we get
\[\int\frac{dv}{v\left( v + 2 \right)} = - \int\frac{dx}{x}\]
\[ \Rightarrow \frac{1}{2}\int\left[ \frac{1}{v} - \frac{1}{v + 2} \right]dv = - \int\frac{dx}{x}\]
\[ \Rightarrow \frac{1}{2}\left[ \int\frac{1}{v}dv - \int\frac{1}{v + 2}dv \right] = - \int\frac{dx}{x}\]
\[ \Rightarrow \frac{1}{2}\left[ \log \left| v \right| - \log \left| v + 2 \right| \right] = - \log \left| x \right| + \log C\]
\[ \Rightarrow \frac{1}{2}\log \left| \frac{v}{v + 2} \right| = \log \left| \frac{C}{x} \right| \]
\[ \Rightarrow \log \left| \frac{v}{v + 2} \right| = 2\log \left| \frac{C}{x} \right|\]
\[ \Rightarrow \log \left| \frac{v}{v + 2} \right| = \log \left| \frac{C^2}{x^2} \right|\]
\[ \Rightarrow \frac{v}{v + 2} = \frac{C^2}{x^2}\]
\[ \Rightarrow \frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C^2}{x^2}\]
\[ \Rightarrow \frac{y}{y + 2x} = \frac{C^2}{x^2}\]
\[ \Rightarrow x^2 y = C^2 \left( y + 2x \right)\]
\[ \Rightarrow x^2 y = K\left( y + 2x \right) ..........\left(\text{Where, }K = C^2 \right)\]