FOR CALCULUS EXPERTS ONLY

[Scrapster]

Check this out:

Integ(exp(au)cos(bu)du = ((exp(au))/(a^2 + b^2)) * (acos(bu) + bsin(bu)) + c

I think this is it.. Hold on. I'll have an answer in a min.

 

[Scrapster]

Oh btw. This was the original eqn.

y' - 1/2y = 2cos(t)

solve for y.

 

[dabuddha]

hmm its been awhile since ive done algebra but wouldnt that be
dy/dt - 1/2*y = 2 cos(t)
dy/dt = 2cost + 1/2y
after the integral
y = 2sint + 1/2y*t
y-1/2yt = 2sint
(1-1/2t)y = 2sint
y = 2sint / (1 - 1/2t)

im sure i screwed up somewhere but how'd you get that other problem from this one
unless i read it wrong and they're two different problems

 

[bonkers325]

math :Q

 

[hendon]

dabuddha:
in your derivation:

dy/dt - 1/2*y = 2 cos(t)
dy/dt = 2cost + 1/2y
after the integral
y = 2sint + 1/2y*t <----- 2nd term on the right doesn't follow fr above
y-1/2yt = 2sint
(1-1/2t)y = 2sint
y = 2sint / (1 - 1/2t)

My take is that you'll have to use the integrating factor method...

 

[Scrapster]

I wish it could be solved like that. Before you get to the part where I'm stuck you have to find the &quot;integrating factor&quot; mu(t) and distribute it. Integrate both sides and solve for y.

 

[dabuddha]

from the way i did it, i integrated with respect to t
hence the integral of 1/2y was 1/2y * t
cause y would be treated as a constant
i could be wrong though bout how i did it

 

[Scrapster]

but in this case y is not a constant.

Makes the problem more interesting.

 

[dabuddha]

i mean y being constant when i integrated
when you integrate with respect to a variable, whatever other variables there are are treated as constants
but for those two problems, thats all i was able to come up with
i double checked what i had

 

[hendon]

Scrapster:

yup.. I think you got the integrating factor part already..
so all you have to do is to integrate the rhs, which is
2*exp(-t/2)*cos(t)
which is done by integration by parts as mentioned by the other people in the thread...

 

[Scrapster]

I used an integration formula that was mentioned already.

Didn't need parts.

 

[dabuddha]

im off to bed
good luck on the problems
let me know when you get them
got me interested in calc again hehe

 

[hendon]

hmm.. then you should be done?
It'll be:

y*exp(-t/2) = value of the integral which you got
then solve for y..

 

[Scrapster]

Hendon,

Yeah, Just got stuck on that integral.