Question

The solution of log $\left[\frac{dy}{dx}\right]=3x+4y$ given that $y\left(0\right)=0$

• A. $4e^{3x}-3e^{-4y}=1$
• B. $3e^{3x}+4e^{4y}=7$
• C. $4e^{3x}+3e^{-4y}=7$
• D. $3e^{3x}+4e^{-y}=1$

Answer 1

Answer: (C)

$4e^{3x}+3e^{-4y}=7$

$log{y}^1=3x+4y$

$\Rightarrow \frac{dy}{dx}=e^{3x+4y}$

$\Rightarrow \int e^{-4y}dy=\int e^{3x}dx+e$

$\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+c$

$3e^{-4y}+4e^{3x}=c$