Question

For the differential equations, find the particular solution satisfying the given condition:$[xsin^2(\frac{x}{y})-y]dx+xdy=0;y=\frac{π}{4},$when $x=1$

Answer 1

$[xsin^2(\frac{x}{y})-y]dx+xdy=0$
$⇒\frac{dy}{dx}=\frac{-[xsin^2(\frac{y}{x})-y]}{x}...(1)$
Let $F(x,y)=\frac{-[xsin^2(\frac{y}{x})-y]}{x}$
$∴F(λx,λy)=[λx.sin^2(\frac{λx}{λy})-λy]=\frac{-[xsin^2(\frac{y}{x})-y]}{x}=λ°.F(x,y)$
Therefore,the given differential equation is a homogenous equation.
To solve this differential equation,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:
$v+x\frac{dv}{dx}=\frac{-[xsin^2v-vx]}{x}$
$⇒v+x\frac{dv}{dx}=-[sin^2v-v]=v-sin^2v$
$⇒x\frac{dv}{dx}=-sin^2v$
$⇒\frac{dv}{sin^2v}=\frac{-dx}{dx}$
$⇒cosec^2\,v\,dv=\frac{-dx}{x}$
$-cotv=-log|x|-C$
$⇒cotv=log|x|+C$
$⇒cot(\frac{y}{x})=log|x|+log|C|$
$⇒cot(\frac{y}{x})=log|Cx|...(2)$
Now,$y=\frac{π}{4}$ at x=1
$⇒cot(\frac{π}{4})=log|C|$
$⇒1=logC$
$⇒C=e^1=e$
Substituting $C=e$ in equation(2),we get:
$cot(\frac{y}{x})=log|ex|$
This is the required solution of the given differential equation.