Question

Solve the following initial value problem:
\[\frac{dy}{dx} = \frac{y\left( x + 2y \right)}{x\left( 2x + y \right)}, y\left( 1 \right) = 2\]

 

Answer 1

\[\frac{dy}{dx} = \frac{y\left( x + 2y \right)}{x\left( 2x + y \right)}, y\left( 1 \right) = 1\]
This is an homogenous equation,

Put `y = vx`

\[\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\Rightarrow v + x\frac{dv}{dx} = \frac{v\left( x + 2vx \right)}{\left( 2x + vx \right)}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v\left( 1 + 2v \right)}{\left( 2 + v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v\left( 1 + 2v \right) - v\left( 2 + v \right)}{\left( 2 + v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v + 2 v^2 - 2v - v^2}{\left( 2 + v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v^2 - v}{\left( 2 + v \right)}\]
\[ \Rightarrow \frac{\left( 2 + v \right)dv}{\left( v^2 - v \right)} = \frac{dx}{x}\]
On integrating both side of the equation we get,
\[\int\frac{2 + v}{\left( v^2 - v \right)}dv = \int\frac{dx}{x}\]
\[ \Rightarrow \int\frac{2}{v\left( v - 1 \right)}dv + \int\frac{v}{v\left( v - 1 \right)}dv = \int\frac{dx}{x}\]
\[ \Rightarrow 2\left[ \int\frac{1}{\left( 1 - v \right)}dv - \int\frac{1}{v}dv \right] + \int\frac{1}{v - 1}dv = \log_e x + c\]
\[ \Rightarrow 2\left[ \log_e \left( v - 1 \right) - \log_e v \right] + \log_e \left( v - 1 \right) = \log_e x + c\]
\[2\left[ \log_e \left( \frac{v - 1}{v} \right) \right] + \log_e \left( v - 1 \right) = \log_e x + c\]
\[2 \log_e \left( \frac{y - x}{y} \right) + \log_e \left( \frac{y - x}{x} \right) = \log_e x + c\]
As `y(1) = 2`
\[2 \log_e \left( \frac{2 - 1}{2} \right) + \log_e \left( \frac{2 - 1}{1} \right) = \log_e 1 + c\]
\[2 \log_e \frac{1}{2} + \log_e 1 = \log_e 1 + c\]
\[ - 2 \log_e 2 + 0 = 0 + c\]
\[ - 2 \log_e 2 = c\]
\[ \therefore 2 \log_e \left( \frac{y - x}{y} \right) + \log_e \left( \frac{y - x}{x} \right) = \log_e x - 2 \log_e 2\]