The particular solution of the differential equation \( \log\left(\displaystyle\frac{dy}{dx}\right)=3x+4y \), given that \( y=0 \), when \( x=0 \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4e^{3x}+3e^{-4y}=7 \)
Since \( \log\left(\frac{dy}{dx}\right)=3x+4y \),
\( \displaystyle \frac{dy}{dx}=e^{3x+4y}=e^{3x}\cdot e^{4y} \)
The variables separate:
\( \displaystyle e^{-4y}\,dy=e^{3x}\,dx \)
\( \displaystyle \int e^{-4y}\,dy=\int e^{3x}\,dx \)
\( \displaystyle -\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}+C \)
At x = 0, y = 0:
\( \displaystyle -\frac{1}{4}=\frac{1}{3}+C \)
\( \displaystyle C=-\frac{7}{12} \)
Multiplying by −12:
\( \displaystyle 3e^{-4y}=-4e^{3x}+7 \)
\( \displaystyle 4e^{3x}+3e^{-4y}=7 \)
Shortcut: Put x = 0, y = 0 into the options: (1) 3 + 4 = 7 ≠ 12, (2) 4 + 3 = 7, (3) 4 + 3 = 7, (4) 3 + 4 = 7 ≠ 12. Only (2) and (3) survive; since \( \frac{dy}{dx}=e^{3x}e^{4y} \) separates into \( e^{-4y}dy=e^{3x}dx \), the answer must contain \( e^{3x} \) and \( e^{-4y} \): Option (3).
NCERT (2026-27 edition):
Methods of Solving First Order, First Degree Differential Equations
