Question

For the differential equations, find the particular solution satisfying the given condition:$2xy+y^2-2x^2\frac{dy}{dx}=0;y=2$ when $x=1$

Answer 1

$2xy+y^2-2x^2\frac{dy}{dx}=0$
$⇒2x^2\frac{dy}{dx}=2xy+y^2$
$⇒\frac{dy}{dx}=\frac{2xy+y^2}{2x^2}...(1)$
Let $F(x,y)=\frac{2xy+y^2}{2x^2}.$
$∴F(λx,λy)=\frac{2(λx)(λy)+(λy)^2}{2(λx)^2}=\frac{2xy+y^2}{2x^2}=λ°.F(x,y)$
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 value of y and $\frac{dy}{dx}$ in equation(1),we get:
$v+x\frac{dv}{dx}=\frac{2x(vx)+(vx)^2}{2x^2}$
$⇒v+x\frac{dv}{dx}=\frac{2v+v^2}{2}$
$⇒v+x\frac{dv}{dx}=\frac{v+v^2}{2}$
$⇒\frac{2}{v^2}dv=\frac{dx}{x}$
Integrating both sides,we get:
$\frac{2.v-2+1}{-2+1}=log|x|+C$
$⇒\frac{-2}{v}=log|x|+C$
$⇒\frac{-2}{\frac{y}{x}}=log|x|+C$
$⇒\frac{-2x}{y}=log|x|+C...(2)$
Now,y=2 at x=1.
$⇒-1=log(1)+C$
$⇒C=-1$
Substituting C=-1 in equation(2),we get:
$\frac{-2x}{y}=log|x|-1$
$⇒\frac{2x}{y}=1-log|x|$
$⇒y=\frac{2x}{1-log|x|},(x≠0,x≠e)$
This is the required solution of the given differential equation.