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# Show that the differential equation is homogeneous and solve $x\;dy-y\;dx=\sqrt{x^2+y^2}dx$

Can you answer this question?

Toolbox:
• A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. • To solve this type of equations substitute y = vx and \large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
We can rewrite the given equation as $\large\frac{dy}{dx }=\large\frac{ \sqrt {x^2 - y^2} + y}{x}$
$F(x,y) =\large\frac{ [\sqrt {x^2+y^2} +y]}{x}$
$F(kx,ky) = \large\frac{\sqrt{ k^2x^2+ k^2y^2} +ky}{kx} =$$k^0.F(x,y) Hence this is a homogenous equation with degree zero. Step 2: Using the information in the tool box let us substitute y = vx and \large\frac{dy}{x} =$$ v + x\large\frac{dv}{dx}$
$v + x\large\frac{dv}{dx} =\frac{ \sqrt{ x^2 + v^2x^2} + vx}{x}$
taking $x$ as the common factorand cancelling we get,
$v + x\large\frac{dv}{dx} $$= \sqrt { 1+v^2)}+v Cancelling v on both sides we get, x\large\frac{dv}{dx}$$= \sqrt{ (1+v^2)}$
Seperating the variables we get,
$\large\frac{dv}{\sqrt{1+v^2}} =\frac{ dx}{x}$
Step 3:
Integrating on both sides we get,
$\log|v+(\sqrt{ 1+v^2})| = \log x + \log C$
writing the value of $v =\large\frac{ y}{x}$
$\log(\large\frac{y}{x}) $$+ \sqrt{ 1+(y^2/x^2) }= \log Cx \log\large\frac{y + \sqrt{(x^2+y^2)}}{x }$$= \log Cx$