Find the particular solution of the differential equation $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ given that $y=0$ when $x=0$.

(N/A) The given differential equation is $\log \left(\frac{d y}{d x}\right)=3 x+4 y$.
By the definition of logarithm,we can write this as $\frac{d y}{d x}=e^{3 x+4 y}$.
This can be expressed as $\frac{d y}{d x}=e^{3 x} \cdot e^{4 y}$.
Separating the variables,we get $\frac{d y}{e^{4 y}}=e^{3 x} d x$,which is $e^{-4 y} d y=e^{3 x} d x$.
Integrating both sides,we get $\int e^{-4 y} d y=\int e^{3 x} d x$.
This gives $\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C$.
Multiplying by $12$,we get $-3 e^{-4 y}=4 e^{3 x}+12 C$,or $4 e^{3 x}+3 e^{-4 y}+K=0$,where $K=12 C$.
Given that $y=0$ when $x=0$,substituting these values into the equation: $4 e^{0}+3 e^{0}+K=0 \implies 4+3+K=0 \implies K=-7$.
Thus,the particular solution is $4 e^{3 x}+3 e^{-4 y}-7=0$.