Question

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

Answer 1

Consider the differential equation,

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

Taking exponent on both the sides, we have

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

`=>dy/dx=e^(3x+4y)`

`=>dy/dx=e^(3x).e^(4y)`

`=>dy/(e^(4y))=e^(3x)dx`

Integration in both the sides, we have

`intdy/e^4y=inte^(3x)dx`

`e^(-4y)/(-4)=e^(3x)/3+C`

We need to find the particular solution.

We have, y=0, when x=0

`1/(-4)=1/3+C`

`=>C=-1/4-1/3`

`=>C=(-3-4)/12=-7/12`

Thus, the solution is `e^(3x)/3+e^(-4y)/4=7/12`