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. $None \,of\, these$

Answer 1

The Correct Option is 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'\left(x\right)dx = \frac{\left(f\left(x\right)\right)^{n+1}}{n+1} $ and $\int\frac{f'\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 $ $\Rightarrow \log\left(xy\right) =\frac{-1}{x+y} +C$