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.