Question

The solution of the differential equation $x^{2}dy+y(x+y)dx=0$ is:

• A. $x^{2}y=c^2(2x+y)$
• B. $x{y}^2=c^2(2x+y)$
• C. $x^{2}y=c^2(x+2y)$
• D. $x{y}^2=c^2(x+2y)$

Answer 1

Answer: (A)

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

Given differential equation can be re-written as follows:
$(\frac{x}{y}{)}^2\frac{dy}{dx}+\frac{x}{y}+1=0$ -------(i)
Substitute $y=xt$

$\Rightarrow \frac{dy}{dx}=t+\frac{dt}{dx}$
With above substitution equation (i) will become:
$\frac{1}{t^2}(t+x\frac{dt}{dx})+\frac{1}{t}+1=0\Rightarrow x\frac{dt}{dx}=-(2t+t^2)\Rightarrow \frac{dt}{t(t+2)}=-\frac{dx}{x}$
$\therefore \frac{1}{2}[\frac{1}{t}-\frac{1}{t+2}]dt=-\frac{dx}{x}$
$\Rightarrow \frac{1}{2}\ln \left(\frac{t}{t+2}\right)=-\ln x+c_1$
$\Rightarrow \ln \left(\frac{t}{t+2}\right)=-2\ln x+\ln k$, where $k=e^{2c_1}$
$\Rightarrow \ln \left(\frac{t}{t+2}\right)=\ln \left(\frac{k}{x^2}\right)$
$\Rightarrow \frac{t}{t+2}=\frac{k}{x^2}$
Back substituting $t=\frac{y}{x}$ in above equation, we will get
$\frac{y}{(2x+y)}=\frac{k}{x^2}$
$\therefore x^{2}y=c^2(2x+y)$, where $k=c^2$