The solution of initial value problem: \( \displaystyle\frac{dy}{dx}=e^{3x+4y};\ y(0)=-\frac{1}{4} \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3e^{-4y}+4e^{3x}=3e+4 \)
Since \( e^{3x+4y}=e^{3x}\cdot e^{4y} \), the variables separate:
\( \displaystyle e^{-4y}\,dy=e^{3x}\,dx \)
\( \displaystyle \int e^{-4y}\,dy=\int e^{3x}\,dx \)
\( \displaystyle -\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}+C \)
Multiplying by −12,
\( \displaystyle 3e^{-4y}=-4e^{3x}-12C \)
\( \displaystyle 3e^{-4y}+4e^{3x}=K\quad(K=-12C) \)
Putting x = 0, y = −1/4, so that \( e^{-4y}=e^{1}=e \):
\( \displaystyle 3e+4=K \)
Hence \( 3e^{-4y}+4e^{3x}=3e+4 \).
Shortcut: Put x = 0, y = −1/4 (so \( e^{-4y}=e \) and \( e^{3x}=1 \)) into each option:
\( \displaystyle \text{Option 1: } e+1\ne 3e \)
\( \displaystyle \text{Option 2: } 3e+4=3e+4 \)
\( \displaystyle \text{Option 3: } -4e+3\ne 3e+4 \)
\( \displaystyle \text{Option 4: } e-3\ne 0 \)
Only Option (2) satisfies the initial condition.