# Show that

Question:

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

Solution:

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$\

$\Rightarrow x d y=\left[y+\sqrt{x^{2}+y^{2}}\right] d x$

$\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$             ...(1)

Let $F(x, y)=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}=\lambda^{0} \cdot F(x, y)$

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 $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+(v x)^{2}}}{x}$

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

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

Integrating both sides, we get:

$\log \left|v+\sqrt{1+v^{2}}\right|=\log |x|+\log \mathrm{C}$

$\Rightarrow \log \left|\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}\right|=\log |\mathrm{C} x|$

$\Rightarrow \log \left|\frac{y+\sqrt{x^{2}+y^{2}}}{x}\right|=\log |\mathrm{C} x|$

$\Rightarrow y+\sqrt{x^{2}+y^{2}}=\mathrm{C} x^{2}$

This is the required solution of the given differential equation.