The general solution of the differential equation $\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{. e^x+e^{-x}}$…
The general solution of the differential equation $\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{. e^x+e^{-x}}$ is
- $y=\mathrm{e}^{-3 x}+\mathrm{c}$, where c is a constant of integration.
- $y=\mathrm{e}^x+\mathrm{c}$, where c is a constant of integration.
- $y=\mathrm{e}^{3 x}+\mathrm{c}$, where c is a constant of integration.
- $y=\mathrm{e}^{-x}+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 \mathrm{e}^{2 x}+3 \mathrm{e}^{4 x}}{\mathrm{e}^x+\mathrm{e}^{-x}} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3 \mathrm{e}^{2 x}\left(1+\mathrm{e}^{2 x}\right)}{\left(\frac{\mathrm{e}^{2 x}+1}{\mathrm{e}^x}\right)} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \mathrm{e}^{2 x} \cdot \mathrm{e}^x \\
& \Rightarrow \mathrm{~d} y=3 \mathrm{e}^{3 x} \mathrm{~d} x
\end{aligned}$
Integrating on both sides, we get
$y=\mathrm{e}^{3 x}+\mathrm{c}$
Asked in: MHT CET 2024 (15 May Shift 2)
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