# Solve the following differential equation :$x\large\frac{dy}{dx}$$-y=\sqrt{x^2+y^2}. ## 1 Answer Toolbox: • A function is said to be homogenous function in degree n if F(kx,ky) = k^nF(x,y) for any non zero constant k. • To solve these type of homogenous functions we make the substitution y = vx hence \large\frac{dy}{dx}$$ = v+x\large\frac{dv}{dx}$
Step 1:
$x\large\frac{dy}{dx}-$$y=\sqrt{x^2+y^2} This can be written as \large\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}-----(1) Step 2: Clearly this is a homogenous differential equation it can solved as follows : Put y=vx On differentiating with respect to x we get \large\frac{dy}{dx}$$=v+x\large\frac{dv}{dx}$
Substituting for $y$ and $\large\frac{dy}{dx}$ in equ (1) we get,
$v+x\large\frac{dv}{dx}=\frac{vx+\sqrt{x^2+v^2x^2}}{x}$
$x\large\frac{dv}{dx}$$=\sqrt{1+v^2} Step 3: Now separating the variables we get, \large\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} Integrating on both sides we get, \int \large\frac{dv}{\sqrt{1+v^2}}=\int\frac{dx}{x} Consider \int \large\frac{dv}{\sqrt{1+v^2}} This is of the form \int \large\frac{dx}{\sqrt{x+a^2}}$$=\log \mid x+\sqrt{x^2+a^2}\mid +c$
$\therefore \log \mid v+\sqrt{1+v^2}=\log x+\log c$
$\Rightarrow \log\mid v+\sqrt{1+v^2}\mid=\log cx$
$\log\mid v+\sqrt{1+v^2}\mid= cx$
Step 4:
Substituting for $v$ we get,
$\large\frac{y}{x}+$$\sqrt{1+\large\frac{y^2}{x^2}}$$=cx$
$\large\frac{y}{x}+$$\large\frac{x^2+y^2}{x}$$=cx$
$\Rightarrow y+x^2+y^2=cx^2$
(i.e)$y^2+y=x^2(c-1)$
Put $c-1=k$
$y^2+y=kx^2$
$kx^2+y^2+y=0$
This is the required solution.