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.