Question

For the given differential equation find the particular solution satisfying the given conditions.

$x^{2}dy+(xy+y^2)dx=0,y=1whenx=1$

Answer 1

Given, $x^{2}dy=-(xy+y^2)dx\Rightarrow \frac{dy}{dx}=-\frac{xy+x^2}{x^2}$
Thus, the given differential equations is homogeneous.
So, put $\frac{y}{x}=vi.e.,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, Eq. (i) becomes $v+x\frac{dv}{dx}=-\frac{x^{2}v+x^2{v}^2}{x^2}\Rightarrow v+x\frac{dv}{dx}=-\frac{x^2(v+v^2)}{x^2}\Rightarrow v+x\frac{dv}{dx}=-(v+v^2)\Rightarrow x\frac{dv}{dx}=-2(2v+v^2)\Rightarrow \frac{dv}{v^2+2v}=-\frac{1}{x}dx$
On integrating both sides, we get $\int \frac{dv}{v^2+2v}=-\int \frac{dx}{x}Letv+1=t\Rightarrow dv=dt\therefore \int \frac{1}{t^2-1}dt=-\int \frac{dx}{x}\Rightarrow \frac{1}{2}log\left|\frac{t-1}{t+1}\right|=-log|x|+C$ $[\because \int \frac{dx}{x^2-a^2}=\frac{1}{2a}log|\frac{x-a}{x+a}+C|]\Rightarrow \frac{1}{2}log\left|\frac{v+1-1}{v+1+1}\right|=-log|x|+logC$ $[\because t=v+1]$
$\Rightarrow \frac{1}{2}log\left|\frac{v}{v+2}\right|+2log|x|=2logC\Rightarrow log\left|\left(\frac{v}{v+2}\right)x^2\right|=log{C}^2\Rightarrow log\left|\left(\frac{\frac{y}{x}}{\frac{y}{x}+2}\right)x^2\right|=log{C}^2(putv=\frac{y}{x})\Rightarrow \frac{x^{2}y}{2x+y}=A$ ...(ii)
[let $C^2=Aandlogm=logn\Rightarrow m=n$]
When x=1, y=1
$\therefore \frac{1}{2+1}=A\Rightarrow A=\frac{1}{3}$
On putting the value of A in Eq. (ii), we get
$\frac{x^{2}y}{2x+y}=\frac{1}{3}\Rightarrow 3x^{2}y=2x+y$
This is the required solution of the given differential equation.