Question

A particular solution of $\log \left(\frac{d y}{d x}\right)=3 x+4 y, y(0)=0$ is

• A. $e^{3 x}+3 e^{-4 y}=4$
• B. $4 e^{3 x}-3 e^{-4 y}=3$
• C. $3 e^{3 x}+4 e^{-4 y}=7$
• D. $4 e^{3 x}+3 e^{-4 y}=7$

Answer 1

Answer: (C)

Given, $\log \left(\frac{d y}{d x}\right)=3 x+4 y$
$\Rightarrow \frac{d y}{d x}=e^{3 x} e^{4 y}$
$\Rightarrow e^{-4 y} d y=e^{3 x} d x$
On integrating both sides, we get
$-4 e^{-4 y}=3 e^{3 x}+c$
At $x=0,\, y=0$,
$-4=3+c \Rightarrow c=-7$
$\therefore $ Solution is $4 e^{-4 y}+3 e^{3 x}=7$