Originally Posted by

**tigerdivision** Hi everyone

I have this particular problem which I would appreciate someone would help me solve:

Find the solution of the initial-value problem:

dy/dx = cos(4x)

------------- , y=6 when x=0

3 - sin(4x)

Here are my workings:

y = ∫ cos (4x) / 3 - sin (4x) dx = ∫ 1/4sin(4x)/3x+1/4cos(4x) dx

y = sin (4x) (3x + cos (4x)) +c since 1/4 cancel each other out

6 = sin (0) (0+cos(0)) + c

6 = 0 + 6

c = 6

I am not sure whether the solution arrived at is correct. Would appreciate some help on this.

Many thanks