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Mathematics Question on integral

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

A

log⁡(xy)=−1x+y+C\log\left(xy\right) =- \frac{1}{x+y} +C

B

log⁡(xy)=−1x+y+C\log\left(\frac{x}{y}\right) =- \frac{1}{x+y} +C

C

log⁡(xy)=−1x−y+C\log\left(xy\right) =- \frac{1}{x - y} +C

D

None of theseNone \,of\, these

Answer

log⁡(xy)=−1x+y+C\log\left(xy\right) =- \frac{1}{x+y} +C

Explanation

Solution

Given D.E is (x+y)2(xdydx+y)=xy(1+dydx)\left(x+y\right)^{2} \left(x \frac{dy}{dx} +y\right) =xy \left(1 + \frac{dy}{dx}\right) ⇒(xy)−1(xdydx+y)=(x+y)−2(1+dydx)\Rightarrow \left(xy\right)^{-1} \left(x \frac{dy}{dx}+y\right)= \left(x+y\right)^{-2} \left(1+\frac{dy}{dx}\right) ⇒∫(xy)−1(xdydx+y)=∫(x+y)−2(1+dydx)\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 ∫(f(x))nf′(x)dx=(f(x))n+1n+1\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 ∫f′(x)f(x)dx=log⁡(f(x))+C\int\frac{f'\left(x\right)}{f\left(x\right)} dx = \log\left(f\left(x\right)\right)+C ⇒log⁡(xy)=(x+y)−1−1+C\Rightarrow \log\left(xy\right) =\frac{\left(x+y\right)^{-1}}{-1} +C ⇒log⁡(xy)=−1x+y+C\Rightarrow \log\left(xy\right) =\frac{-1}{x+y} +C