Thread: Some help required..

1. Some help required..

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

2. 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
Ok, so i dont really get what you were doing, but here's the way i'd do this.

we have dy/dx = cos(4x)/(3 - sin(4x))

=> dy = cos(4x)/(3 - sin(4x)) dx
=> y = int{cos(4x)/(3 - sin(4x))}dx
we proceed by substitution:
Let u = 3 - sin(4x)
=> du = -4cos(4x) dx
=> (-1/4) du = cos(4x) dx
so our integral becomes:

y = (-1/4)*int{1/u}du
= (-1/4)lnu + C
= (-1/4)ln(3 - sin(4x)) + C

when x = 0, y = 6
=> 6 = (-1/4)ln(3) + C
so C = 6 + (ln3)/4

so y = (-1/4)ln(3 - sin(4x)) + 6 + (ln3)/4 is our solution

3. 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
oh, i think i see what you were doing now. You thought the integral of a rational function was just the integral of the top divided by the integral of the bottom. I'm afraid that's not so. The above method is the simplest way i see to do this problem