Question

The solution of $\frac{dy}{dx}=e^{3x+4y}$ with $y\left(0\right)=0$ is $4e^{3x}+3e^{-4y}=\alpha ,$ then the value of $\alpha$ is

Answer 1

$\frac{dy}{dx}=e^{3x}{e}^{4y}$
$\Rightarrow \int e^{-4y}dy=\int e^{3x}dx$
$\Rightarrow \frac{-e^{-4y}}{4}=\frac{e^{3x}}{3}+C$
$\because y\left(0\right)=0$
$\therefore C=\frac{-1}{4}-\frac{1}{3}=\frac{-7}{12}$

Hence, $4e^{3x}+3e^{-4y}=7$
$\Rightarrow \alpha =7$