Question

The solution of the differential equation $\frac{dy}{dx}=\frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}$is

• A. $y=e^{3x}+C$
• B. $y=2e^{2x}+C$
• C. $y=e^x+C$
• D. $y=e^{4x}+C$
• E. $y=9e^{3x}+C$

Answer 1

Answer: (A)

Given differential equation is
$\frac{dy}{dx}=\frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}=\frac{3e^{2x}\left(1+e^{2x}\right)\cdot e^x}{\left(1+e^{2x}\right)}$
$\Rightarrow \frac{dy}{dx}=3\cdot e^{3x}$
$\Rightarrow \int dy=\int 3\cdot e^{3x}dx$ (on integrating)
$\Rightarrow y=3\cdot \frac{e^{3x}}{3}+C$
$\Rightarrow y=e^{3x}+C$