Question

$log\left(\frac{dy}{dx}\right)=3x+4y;y=0$ when $x=0$.

Answer 1

$log\left(\frac{dy}{dx}\right)=3x+4y$

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

$\frac{dy}{e^{4y}}=e^{dx}dx$

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

applying $y\left(0\right)=0$, we get

$C=\frac{-7}{12}$

So, $\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}-\frac{7}{12}$