Solution of the differential equation e^x-e^ -x e^x+e^ -x =d x-d … - Vidyadip

MCQ

Solution of the differential equation $\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{d x-d y}{d x+d y}$, is

A
$2 y e^{2 x}=C \cdot e^{2 x}+1$
✓
$2 y e^{2 x}=C \cdot e^{2 x}-1$
C
$y e^{2 x}=C \cdot e^{2 x}+2$
D
none of these

✓

Answer

Correct option: B.

$2 y e^{2 x}=C \cdot e^{2 x}-1$

(b): Given equation $\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{d x-d y}{d x+d y}$
Applying componendo and dividendo, we get
$
\begin{array}{l}
\frac{d y}{d x}=\frac{e^{-x}}{e^x} \Rightarrow d y=e^{-2 x} d x \Rightarrow 2 y=-e^{-2 x}+C (Integrating both sides) \\
\Rightarrow 2 y e^{2 x}=C \cdot e^{2 x}-1 \\
\end{array}
$

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