The particular solution of the differential equation \(\log\left(\displaystyle\frac{dy}{dx}\right)=3x+4y\) satisfying \(y = 0\) when \(x = 0\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4e^{3x}+3e^{-4y}-7=0\)
Removing the logarithm, the variables separate:
\( \displaystyle \frac{dy}{dx} = e^{3x+4y} = e^{3x}\,e^{4y} \)
\(e^{-4y}\,dy = e^{3x}\,dx\)
Integrating both sides:
\( \displaystyle -\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} + C \)
Putting x = 0, y = 0:
\( \displaystyle -\frac{1}{4} = \frac{1}{3} + C,\quad C = -\frac{7}{12} \)
Multiplying the solution by −12:
\(3e^{-4y} = -4e^{3x} + 7\)
\(4e^{3x}+3e^{-4y}-7=0\)
Shortcut: The particular solution must be satisfied by x = 0, y = 0, where \(e^{3x} = e^{-4y} = 1\). The left sides of the options become 3 + 4 + 7 = 14, 3 − 4 − 7 = −8, 4 + 3 − 7 = 0 and 4 − 3 + 7 = 8. Only Option (3) gives 0.
NCERT (2026-27 edition):
Differential Equations
