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MHT CET · Maths · Differential Equations

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

  1. A \(y=\mathrm{e}^{-3 x}+\mathrm{c}\), where c is a constant of integration.
  2. B \(y=\mathrm{e}^x+\mathrm{c}\), where c is a constant of integration.
  3. C \(y=\mathrm{e}^{3 x}+\mathrm{c}\), where c is a constant of integration.
  4. D \(y=\mathrm{e}^{-x}+\mathrm{c}\), where c is a constant of integration.
Verified Solution

Answer & Solution

Correct Answer

(C) \(y=\mathrm{e}^{3 x}+\mathrm{c}\), where c is a constant of integration.

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 \mathrm{e}^{2 x}+3 \mathrm{e}^{4 x}}{\mathrm{e}^x+\mathrm{e}^{-x}} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{3 \mathrm{e}^{2 x}\left(1+\mathrm{e}^{2 x}\right)}{\left(\frac{\mathrm{e}^{2 x}+1}{\mathrm{e}^x}\right)} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \mathrm{e}^{2 x} \cdot \mathrm{e}^x \\
& \Rightarrow \mathrm{~d} y=3 \mathrm{e}^{3 x} \mathrm{~d} x
\end{aligned}\)
Integrating on both sides, we get
\(y=\mathrm{e}^{3 x}+\mathrm{c}\)