The solution of ( x + y )^2 d y d x =1 is....... - Vidyadip

MCQ

The solution of $( x + y )^2 \frac{d y}{d x}=1$ is.......

A
$x=\tan ^{-1}(x+y)+c$
B
$y \tan ^{-1}\left(\frac{x}{y}\right)= c$
✓
$y=\tan ^{-1}(x+y)+c$
D
$y+\tan ^{-1}(x+y)=c$

✓

Answer

Correct option: C.

$y=\tan ^{-1}(x+y)+c$

$y=\tan ^{-1}(x+y)+c$
Hint:
$(x+y)^2 \frac{d y}{d x}=1$
Put $x+y=u \quad \therefore 1+\frac{d y}{d x}=\frac{d u}{d x}$
$\therefore u^2\left(\frac{d u}{d x}-1\right)=1$
$\therefore u^2 \frac{d u}{d x}=u^2+1$
$\therefore \int \frac{u^2}{u^2+1} d u=\int d x$
$\therefore \int\left(1-\frac{1}{u^2+1}\right) d u=\int d x$
$\therefore u-\tan ^{-1} u=x+c $
$\therefore x+y-\tan ^{-1}(x+y)=x+c $
$\therefore y=\tan ^{-1}(x+y)+c$

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