Question

For each of the differential equations in Exercises from $11$ to $15$, find the particular solutions satisfying the given condition:
$x^{2}dy+(xy+y^2)dx=0;y=1$ when $x=1$

Answer 1

The differential eq. can be

$x^{2}dy=-(xy+y^2)dx$

$\frac{dy}{dx}=\frac{-(xy+y^2)}{x^2}$ $--------\left(1\right)$

Let $f(x,y)=\frac{dy}{dx}=\frac{-(xy+y^2)}{x^2}$

$f(\lambda x,\lambda y)$ finding

$f(\lambda x,\lambda y)=\frac{-(\lambda x\lambda y+{\lambda }^2{y}^2)}{{\lambda }^2{x}^2}$

$=\frac{-{\lambda }^2(xy+y^2)}{{\lambda }^2{x}^2}=\frac{-(xy+y^2)}{x^2}$

$={\lambda }^{o}f(x,y)$

$=F(x,y)=\frac{-(xy+y^2)}{x^2}$

$\therefore F(x,y)$ is a homogeneous function of degree zero

putting $y=vx$.

Differentiate w.r.t ${}^{\prime }{x}^{\prime }$

$\frac{dy}{dx}=x\frac{dv}{dx}+v$

Putting value of $\frac{dy}{dx}$ and $y=vx$ in $\left(1\right)$

$\frac{dy}{dx}=\frac{-(xy+y^2)}{x^2}$

$v+\frac{dv}{dx}=\frac{-\left(x\right(vx)+(vx{)}^2)}{x^2}$

$v+\frac{xdv}{dx}=\frac{-(x^{2}v+x^2{v}^2)}{x^2}$

$v+\frac{xdv}{dx}=-x^2\frac{(v+v^2)}{x^2}$

$v+x\frac{dv}{dx}=-(v^2+v)$

$\frac{xdv}{dx}=-v^2-v-v$

$\frac{xdv}{dx}=-v^2-2v$

$\frac{dv}{v^2+2v}=\frac{-dx}{x}$

$\int \frac{dv}{v^2+2v}+-\int \frac{dx}{x}$

$\int \frac{dv}{v^2+2v+1-1}=-\log x+\log e$

$\frac{1}{2}\log \frac{v+1-1}{v+1+1}=-\log x+\log e$

$\log \frac{\sqrt{v}}{\sqrt{v+2}}=-\log x+\log e$

$\log \frac{x\sqrt{v}}{\sqrt{v+2}}=\log e$

$\frac{x\sqrt{v}}{\sqrt{v+2}}=c.[v=\frac{y}{x}]$

$\frac{x\sqrt{\frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c=\frac{\sqrt{xy}}{\sqrt{\frac{y+2x}{x}}}$

$\frac{x\sqrt{y}}{\sqrt{y+2x}}=c$

$x\sqrt{y}=c\sqrt{y+2x}$

$S.O.B.S$

$x^{2}y=c^2(y+2x)-------\left(2\right)$

Put $x=1$ & $y=1$

$1^2.1=c^2(1+2)$

$1=3c^2$

$c^2=\frac{1}{3}$

Put $c^2$ in $\left(1\right)$

$x^{2}y=\frac{1}{3}(y+2x)$

$(y+2x)=x^{2}y$.