Question

For the differential equations, find the particular solution satisfying the given condition:$\frac{dy}{dx}-\frac{y}{x}+cosec(\frac{y}{x})=0;y=0$ when $x=1$

Answer 1

$\frac{dy}{dx}-\frac{y}{x}+cosec(\frac{y}{x})=0$
$⇒\frac{dy}{dx}=\frac{y}{x}-cosec(\frac{y}{x})...(1)$
Let $F(x,y)=\frac{y}{x}-cosec(\frac{y}{x}).$
$∴F(λx,λy)=\frac{λy}{λx}-cosec(\frac{λy}{λx})$
⇒F(λx,λy)=y/x-cosec(y/x)=F(x,y)=λ°.F(x,y)
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
$y=vx$
$⇒\frac{d}{dx}(y)=\frac{d}{dx}(vx)$
$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
Substituting the values of y and $\frac{dy}{dx}$ in equation(1),we get:
$v+x\frac{dv}{dx}=v-cosec\,v$
$⇒\frac{-dv}{cosec\,v}=\frac{-dx}{x}$
Integrating both sides we get:
$cos\,v=logx+logC=log|Cx|$
$⇒cos(\frac{y}{x})=log|Cx|...(2)$
This is the required solution of the given differential equation.
Now,y=0 at x=1.
$⇒cos(0)=logC$
$⇒1=logC$
$⇒C=e^1=e$
Substituting C=e in equation(2),we get:
$cos(\frac{y}{x})=log|(ex)|$
This is the required solution of the given differential equation.