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The solution of the differential equation $\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}}$ |
$y=\left(e^x+e^{-x}\right)+C$ $y=Ce^x$ $y=e^{3 x}+C$ $y=3 e^{3 x}+C$ |
$y=e^{3 x}+C$ |
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 $\int dy=\int 3 e^{3 x}dx$ $⇒y=e^{3 x}+c$ |
