Differential Equations - 1

Prob: Find the solution of the differential equation

log (dy/dx) = 3x + 4y, y(0) = 0


Solution:

log (dy/dx) = 3x + 4y

=> dy/dx = e^3x+4y = e^3xe^4y

=> e^-4ydy = e^3xdx

Integrating both sides

e^-4y/-4 = e^3x/3 +C

Using the condition y(0) = 0

-1/4 = 1/3 +C

=> C = -1/4 - 1/3 = -7/12

=>e^-4y/-4 = e^3x/3 -7/12

=> -(12/4)e^-4y = (12/3)e^3x - 7

=>3e^-4y + 4e^3x = 7