Question

For the differential equations, find the particular solution satisfying the given condition:\(x^2dy+(xy+y^2)dx=0;y=1\) when \(x=1\)

Answer 1

\(x^2dy+(xy+y^2)dx=0\)
\(⇒x^2dy=-(xy+y^2)dx\)
\(⇒\frac{dy}{dx}=\frac{-(xy+y^2)}{x^2}...(1)\)
Let \(F(x,y)=\frac{-(xy+y^2)}{x^2}.\)
\(∴F(λx,λy)=\frac{[λx.λy+(λy)^2]}{(λx)^2}=\frac{-(xy+y^2)}{x}=λ°.F(x,y)\)
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:
\(v+x\frac{dv}{dx}=\frac{-[x.vx+(vx)^2]}{x^2}=-v-v^2\)
\(⇒x\frac{dv}{dx}=-v^2-2v=-v(v+2)\)
\(⇒\frac{dv}{v(v+2)}=\frac{-dx}{x}\)
\(⇒\frac{1}{2}[\frac{(v+2)-v}{v(v+2)}]dv=\frac{-dx}{x}\)
\(⇒\frac{1}{2}[\frac{1}{v}-\frac{1}{v+2}]dv=\frac{-dx}{x}\)
Integrating both sides,we get:
\(\frac{1}{2}[logv-log(v+2)]=-logx+logC\)
\(⇒\frac{1}{2}log(\frac{v}{v+2})=log\frac{C}{x}\)
\(⇒\frac{v}{v+2}=(\frac{c}{x})^2\)
\(⇒\frac{\frac{y}{x}}{\frac{y}{x}+2}=(\frac{C}{x})^2\)
\(⇒\frac{y}{y+2x}=\frac{C^2}{x^2}\)
\(⇒\frac{x^2y}{y^2+2x}=C^2...(2)\)
Now,y=1 at x=1.
\(⇒\frac{1}{1+2}=C^2\)
\(⇒C^2=\frac{1}{3}\)
Substituting \(C^2=\frac{1}{3}\) in equation(2),we get:
\(\frac{x^2y}{y+2x}=\frac{1}{3}\)
\(⇒y+2x=3x2y\)
This is the required solution of the given differential equation.