# Thread: initial-value problem and integrating..... my brain just isn't grasping help

1. ## initial-value problem and integrating..... my brain just isn't grasping help

Hi, i have this initial-value problem....i need to use the reverse quotient rule, but i also need to use an integration formula too:

dy/dx= (cos(5x))/(7+sin(5x)) when y=6 when x=0

f'(x)/f(x)dx=in(f(x))+c (f(x)>0)

i don't want the answer i would just like to know how??? if that makes sense lol. I cannot find an example that is similar to this so i can work through it as i find this is the best way of learning....can anyone help pretty please??

2. ## Re: initial-value problem and integrating..... my brain just isn't grasping help

Use a direct substitution by let u(x) = 7 + sin(5x). Then du = 5cos(5x) dx. Then (1/5) du = cos(5x) dx. Then make the replacements.

dy/dx = (cos (5x)) / (7 + sin (5x))

y = INT {cos (5x) / (7 + sin (5x))} dx * I don't know how to make the integral sign on here *
y = INT {(1/5) 1/u} du
y = (1/5) ln |u| + C
y = (1/5) ln |7 + sin(5x)| + C

To find C, let y = 6 and x = 0.

6 = (1/5) ln |7 + sin(5*0)| + C
6 = (1/5) ln |7 + 0| + C
6 = (1/5) ln 7 + C
6 - 1/5 ln 7 = C

Then y = (1/5) ln |7 + sin(5x)| + 6 - 1/5 ln 7

y = (1/5) ln |(7 + sin(5x))/7| + 6
y = (1/5) ln |1 + (1/7)sin(5x)| + 6

As for the second question, simply let y = f(x) and so then dy/dx = f ' (x). Then you see that INT {f'(x) / f(x)} dx = INT {1/y} dy = ln |y| + C = ln |f(x)| + C. Since f(x) > 0 we see that |f(x)| = f(x). Thus, we see the answer is ln (f(x)) + C as needed.