Question

The general solution of $\frac{dy}{dx}-\frac{y}{x}=\frac{y^2}{x^2}$ is

• A. $\frac{x}{y}=-ln{x}^2+C$
• B. $\frac{x}{y}=-ln\frac{x}{2}+C$
• C. $\frac{x}{y}=-lnx+C$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{x}{y}=-lnx+C$

$\frac{dy}{dx}\frac{-y}{x}=\frac{y^2}{x^2}$ Isn’t this equation of the same form as $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y^{n}withP\left(x\right)=\frac{-1}{x}Q\left(x\right)-=\frac{1}{x^2}andn=2$
This we know is the form of differential equation and we proceed by dividing by $y^n$. i.e., $y^2$ in this case.
Dividing both sides by $y^2$ we get.
$\frac{1}{y^2}\frac{dy}{dx}\frac{-1}{xy}=\frac{1}{x^2}$
Put $y^{-n+1}=1i.e.,\frac{1}{y}=t$
$\frac{-1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$
The given differential equation becomes
$\frac{-dt}{dx}\frac{-t}{x}=\frac{1}{x^2}$
$\frac{dt}{dx}+\frac{t}{x}=\frac{-1}{x^2}$. which is a linear differential equation in $\frac{dt}{dx}$.
I.F.$=e^{\int \frac{1}{x}dx}=e^{lnx}=x$.
General solution is
$tx=\int \frac{-1}{x^2}xdx+Ctx=-lnx+C.\frac{x}{y}=-lnx+C.$