dy/dx=e^(3x+4y)
Let 3x+4y=z
⇒3+4 dy/dx=dz/dx
⇒dy/dx=(dz/dx-3)/4
Substituting the value of dy/dx
(dz/dx-3)/4=e^z
⇒dz/dx=4e^z+3
⇒dz/(4e^z+3)=dx
IIntegrating both the sides
Int ( e^z.dz)/(4(e^z)^2+3e^z)=Int dx
Put e^z=t ⇒e^z.dz=dt
Int dt/(4t^2+3t)=x+c
1/4. Int dt/(t+3/8)^2-9/64)=x+c
⇒ 1/4 4/3 log(4t-3)/(4t+3)=x+c
⇒1/3 log(4e^3x+4y-3)/(4e^3x+4y+3)=x+c
Given x=0 and y=0
⇒c=1/3 log(1/7)
Substitute the value of c and the particular solution is
1/3 { log7(4e^3x+4y-3)/(4e^3x+4y+3)}=x