Thank you for reporting, we will resolve it shortly
Q.
The particular solution of $\log \frac{ dy }{ dx }=3 x +4 y , y (0)=0$ is
Differential Equations
Solution:
$\frac{ dy }{ dx }= e ^{3 x +4 y }= e ^{3 x } \cdot e ^{4 y }$
$\Rightarrow e ^{-4 y } dy = e ^{3 x } dx $
$\Rightarrow \frac{ e ^{4 y }}{-4}=\frac{ e ^{3 x }}{3}+ c$
put $x =0$
We have $-\frac{1}{4}-\frac{1}{3}= c $
$\Rightarrow c =-\frac{7}{12}$,
$\therefore \frac{ e ^{-4 y }}{-4}=\frac{ e ^{3 x }}{3}-\frac{7}{12}$
$ \Rightarrow 7=3 e ^{-4 y }+4 e ^{3 x }$