Question

\[\frac{dy}{dx} = \frac{x}{2y + x}\]

Answer 1

We have,
\[\frac{dy}{dx} = \frac{x}{2y + x}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx \text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x}{2vx + x}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1}{2v + 1}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1}{2v + 1} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - 2 v^2 - v}{2v + 1}\]
\[ \Rightarrow \frac{2v + 1}{1 - 2 v^2 - v}dv = \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int\frac{2v + 1}{1 - 2 v^2 - v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{2v + 1}{2 v^2 + v - 1}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{2v + 1}{2v\left( v + 1 \right) - 1\left( v + 1 \right)}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{2v + 1}{\left( 2v - 1 \right)\left( v + 1 \right)}dv = - \int\frac{1}{x}dx . . . . . (1)\]
Solving left hand side integral of (1), we get
Using partial fraction,
\[\text{ Let }\frac{2v + 1}{\left( 2v - 1 \right)\left( v + 1 \right)} = \frac{A}{\left( 2v - 1 \right)} + \frac{B}{\left( v + 1 \right)}\]
\[ \therefore A + 2B = 2 . . . . . (2) \]
And A - B = 1 . . . . . (3) 
Solving (2) and (3), we get 
\[A = \frac{4}{3}\text{ and }B = \frac{1}{3}\]
\[ \therefore \int\frac{2v + 1}{\left( 2v - 1 \right)\left( v + 1 \right)}dv = \frac{4}{3}\int\frac{1}{2v - 1}dv + \frac{1}{3}\int\frac{1}{v + 1}dv\]
\[ = \frac{4}{3 \times 2}\log \left| 2v - 1 \right| + \frac{1}{3}\log \left| v + 1 \right| + \log C \]
From (1), we get
\[ \frac{2}{3}\log \left| 2v - 1 \right| + \frac{1}{3}\left| v + 1 \right| + \log C = - \log \left| x \right| + \log C_1 \]
\[ \Rightarrow \log \left\{ \left| \left( 2v - 1 \right)^2 \right|\left| v + 1 \right| \right\} = - 3\log\left| x \right| + \log C_2 \]
\[ \Rightarrow \log \left\{ \left| \left( 2v - 1 \right)^2 \right|\left| v + 1 \right| \right\} = \log \left| \frac{{C_2}^3}{x^3} \right|\]
\[ \Rightarrow \left( 2v - 1 \right)^2 \left( v + 1 \right) = \frac{{C_2}^3}{x^3}\]
\[\text{Putting }v = \frac{y}{x},\text{we get }\]
\[ \Rightarrow \left( \frac{2y - x}{x} \right)^2 \left( \frac{y + x}{x} \right) = \frac{{C_2}^3}{x^3}\]
\[ \Rightarrow \left( x + y \right) \left( 2y - x \right)^2 = k\]