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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve the following differential equation :$x\large\frac{dy}{dx}$$-y=\sqrt{x^2+y^2}$.

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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.
answered Sep 20, 2013 by sreemathi.v
 

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