Find the particular solution of the differential equation log(dy/dx) = 3x + 4y, given that y = 0 when x = 0.

Let I = log(dy/dx) = 3x + 4y
→ dy/dx = e^(3x).e^(4y)
→ e^(-4y)dy = e^(3x)dx
→ ∫e^(-4y)dy = ∫e^(3x)dx
→ e^(-4y)/(-4) = e^(3x)/3 + C
Putting y = 0, x = 0
C = -1/4
Therefore, the required solution is
e^(-4y)/(-4) = e^(3x)/3 - 1/4
i.e., 4e^(3x) + 3e^(-4y) = 7