Find the particular solution of the differential equation $\frac{d y}{d x}=\frac{y(2 y-x)}{x(2 y+x)}$ given that $y=1$ when $x=1$.

Find the particular solution of the differential equation $\frac{d y}{d x}=\frac{y(2 y-x)}{x(2 y+x)}$ given that $y=1$ when $x=1$ .

Find the particular solution of the differential equation $\frac{d y}{d x}=\frac{y(2 y-x)}{x(2 y+x)}$ given that $y=1$ when $x=1$ .

Solution:

$\begin{array}{l} \Rightarrow \frac{d y}{d x}=\frac{y(2 y-x)}{x(2 y+x)} \\ \Rightarrow \frac{d y}{d x}=\frac{y\left(2^{y}-1\right)}{x\left(2_{x}^{\frac{y}{x}}+1\right)} \\ \Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right) \end{array}$
$\Rightarrow$ the given differential eq. is a homogenous eq.
Solution of the given differential eq. is:
$\begin{array}{l} \text { Put } y=v x \\ \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ v+x \frac{d v}{d x}=\frac{v x\left(2 \frac{v x}{x}-1\right)}{x\left(2 \frac{v x}{x}+1\right)}=v\left(\frac{2 v-1}{2 v+1}\right) \\ \Rightarrow x \frac{d v}{d x}=v\left(\frac{2 v-1}{2 v+1}\right)-v \\ \Rightarrow x \frac{d v}{d x}=v\left(\frac{2 v-1-2 v-1}{2 v+1}\right) \Rightarrow x \frac{d v}{d x}=\frac{-2 v}{2 v+1} \\ \Rightarrow \frac{2 v+1}{2 v} d v=\frac{-d x}{x} \\ \Rightarrow d v+\left(\frac{1}{2 v}\right) d v=\frac{-d x}{x} \end{array}$
On integrating both sides we obtain:
$\begin{array}{l} \Rightarrow \int\left(\mathrm{dv}+\left(\frac{1}{2 \mathrm{v}}\right) \mathrm{dv}\right)=-\int \frac{\mathrm{dx}}{\mathrm{x}}+c \\ \Rightarrow \mathrm{v}+\frac{\ln |\mathrm{v}|}{2}=-\ln |\mathrm{x}|+c \end{array}$
By resubstituting the value of $y=v x$ we have
$\begin{array}{l} \Rightarrow \frac{y}{x}+\frac{\ln \left|\frac{y}{x}\right|}{2}=-\ln |x|+c \\ y=1 \text { when } x=1 \\ 1+0=-0+c \end{array}$
$\begin{array}{l} \Rightarrow c=1 \\ \Rightarrow \frac{y}{x}+\frac{1}{2} \log |x y|=1 \end{array}$