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
$\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^{x}+e^{-x}}=\frac{3 e^{2 x}\left(1+e^{2 x}\right) \cdot e^{x}}{\left(1+e^{2 x}\right)}$
$\Rightarrow \, \frac{d y}{d x}=3 \cdot e^{3 x}$
$\Rightarrow \, \int d y=\int 3 \cdot e^{3 x} d x \quad$ (on integrating)
$\Rightarrow \,y=3 \cdot \frac{e^{3 x}}{3}+C$
$\Rightarrow \, y=e^{3 x}+C$