The solution of the differential equation $\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}}$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $y=e^{3 x}+C$
$\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}}=\frac{3e^{3x}(e^{-x}+e^x)}{e^x+e^{-x}}$
so $\int dy=\int 3 e^{3 x}dx$
$⇒y=e^{3 x}+c$