# Show that

Question:

$(x+y) d y+(x-y) d x=0 ; y=1$ when $x=1$

Solution:

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

$\Rightarrow(x+y) d y=-(x-y) d x$

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

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

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

Therefore, 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:

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

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

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

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

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

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

Integrating both sides, we get:

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

$\Rightarrow \log \left(1+v^{2}\right)+2 \tan ^{-1} v=-2 \log x+2 k$

$\Rightarrow \log \left[\left(1+v^{2}\right) \cdot x^{2}\right]+2 \tan ^{-1} v=2 k$

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

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

Now, $y=1$ at $x=1$.

$\Rightarrow \log 2+2 \tan ^{-1} 1=2 k$

$\Rightarrow \log 2+2 \times \frac{\pi}{4}=2 k$

$\Rightarrow \frac{\pi}{2}+\log 2=2 k$

Substituting the value of $2 k$ in equation (2), we get:

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

This is the required solution of the given differential equation.