Question

The solution of ${\left(x+y\right)}^2\left(x\frac{dy}{dx}+y\right)=xy\left(1+\frac{dy}{dx}\right)$ is

• A. $\log \left(xy\right)=-\frac{1}{x+y}+C$
• B. $\log \left(\frac{x}{y}\right)=-\frac{1}{x+y}+C$
• C. $\log \left(xy\right)=-\frac{1}{x-y}+C$
• D. $Noneofthese$

Answer 1

Answer: (A)

Given D.E is
${\left(x+y\right)}^2\left(x\frac{dy}{dx}+y\right)=xy\left(1+\frac{dy}{dx}\right)$
$\Rightarrow {\left(xy\right)}^{-1}\left(x\frac{dy}{dx}+y\right)={\left(x+y\right)}^{-2}\left(1+\frac{dy}{dx}\right)$
$\Rightarrow \int {\left(xy\right)}^{-1}\left(x\frac{dy}{dx}+y\right)=\int {\left(x+y\right)}^{-2}\left(1+\frac{dy}{dx}\right)$
Using integral $\int {\left(f\left(x\right)\right)}^n{f}^{\prime }\left(x\right)dx=\frac{{\left(f\left(x\right)\right)}^{n+1}}{n+1}$ and $\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log \left(f\left(x\right)\right)+C$
$\Rightarrow \log \left(xy\right)=\frac{{\left(x+y\right)}^{-1}}{-1}+C$
$\Rightarrow \log \left(xy\right)=\frac{-1}{x+y}+C$