General solution of the differential equation \( (e^x+e^{-x})dy=(3e^{2x}+3e^{4x})\,dx \) is :
[Where C is an arbitrary constant]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( y=e^{3x}+C \)
The variables are already separate, and the right side simplifies before integrating.
\( \displaystyle \frac{dy}{dx}=\frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}} \)
Taking \( 3e^{2x} \) out of the numerator and \( e^{-x} \) out of the denominator:
\( \displaystyle \frac{dy}{dx}=\frac{3e^{2x}(1+e^{2x})}{e^{-x}(e^{2x}+1)} \)
\( \displaystyle =3e^{2x}\cdot e^{x} \)
\( \displaystyle =3e^{3x} \)
Integrating,
\( \displaystyle y=\int 3e^{3x}\,dx \)
\( \displaystyle =3\cdot\frac{e^{3x}}{3}+C \)
\( \displaystyle =e^{3x}+C \)
Why the other options are incorrect
Option (1) \( y=e^x+C \) — Incorrect. The derivative of \( e^x \) is \( e^x \), not \( 3e^{3x} \).
Option (2) \( y=3e^x+C \) — Incorrect. The derivative of \( 3e^x \) is \( 3e^x \), not \( 3e^{3x} \).
Option (4) \( y=3e^{3x}+C \) — Incorrect. This forgets to divide by 3 when integrating \( e^{3x} \); the derivative of \( 3e^{3x} \) is \( 9e^{3x} \).