The solution of the equation d y d x = e^x+e^ -x e^x-e^ -x is : - Vidyadip

MCQ

The solution of the equation $\frac{d y}{d x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}$ is :

A
$y=\log \left(e^x+e^{-x}\right)+c$
✓
$y=\log \left(e^x-e^{-x}\right)+c$
C
$y=\log \left(e^x+1\right)+c$
D
$y=\log \left(1-e^{-x}\right)+c$

✓

Answer

Correct option: B.

$y=\log \left(e^x-e^{-x}\right)+c$

(B)
$
\frac{d y}{d x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}
$
Separating the variables
$
d y=\frac{e^x+e^{-x}}{e^x-e^{-x}} d x
$
Hence $\quad \int d y=\int \frac{e^x+e^{-x}}{e^x-e^{-x}} d x$ $
\Rightarrow \quad y=\log \left(e^x-e^{-x}\right)+c
$
Hence the correct choice is (B).

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