# Show that

Question:

$(x-y) d y-(x+y) d x=0$

Solution:

The given differential equation is:

$(x-y) d y-(x+y) d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x-y}$             ..(1)

$\operatorname{Let} F(x, y)=\frac{x+y}{x-y}$

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}=\lambda^{0} \cdot F(x, y)$

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v}$

$x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1-v}$

$\Rightarrow \frac{1-v}{\left(1+v^{2}\right)} d v=\frac{d x}{x}$

$\Rightarrow\left(\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}}\right) d v=\frac{d x}{x}$

Integrating both sides, we get:

$\tan ^{-1} v-\frac{1}{2} \log \left(1+v^{2}\right)=\log x+\mathrm{C}$

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left[1+\left(\frac{y}{x}\right)^{2}\right]=\log x+\mathrm{C}$

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)=\log x+\mathrm{C}$

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\left[\log \left(x^{2}+y^{2}\right)-\log x^{2}\right]=\log x+\mathrm{C}$

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \log \left(x^{2}+y^{2}\right)+\mathrm{C}$

This is the required solution of the given differential equation.