ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: berryy on June 01, 2011, 07:25:50 pm

Title: integrating by parts
Post by: berryy on June 01, 2011, 07:25:50 pm: can someone please show me how to solve for q via integrating by parts for:

dq/dt +10q = .25sin5t

i keep getting my constants wrong but when i do it exponetially it works,

Thanks all
Title: Re: integrating by parts
Post by: moekamo on June 01, 2011, 08:06:14 pm: im pretty sure you cant do that with integration by parts...

you can do it though, first find the solution to the homogeneous equation:

$\frac{dq}{dt}+10q=0$ - this is seperable, and should be easy ( $q_h=C\cdot e^{-10t}$ )

then for the particular solution, you take a guess that
$q_p=A\sin(5t)+B\cos(5t)$

finding $\frac{dq}{dt}$ and subbing back in, then solving for A, B gives the particular solution ( $q_p=\frac{1}{50} \sin 5t - \frac{1}{100} \cos 5t$ )

so the overall, general solution is:

$q(t)=q_h+q_p=C\cdot e^{-10t} + \frac{1}{50} \sin 5t - \frac{1}{100} \cos 5t$
Title: Re: integrating by parts
Post by: jimmy999 on June 01, 2011, 09:04:05 pm: You can also do that question using an integration factor. I'm pretty sure by using that method you'll then have to use integration by parts. I won't actually go into it, but a hint, you'll have to use integration by parts twice and collect the common factors
Title: Re: integrating by parts
Post by: berryy on June 01, 2011, 09:30:37 pm: Quote from: jimmy999 on June 01, 2011, 09:04:05 pm
You can also do that question using an integration factor. I'm pretty sure by using that method you'll then have to use integration by parts. I won't actually go into it, but a hint, you'll have to use integration by parts twice and collect the common factors

yeah thats what i meant!
when i integrated it twice, ive stuffed something up when i rearranged the "I" cos my constants are not 1/50 or 1/100
Title: Re: integrating by parts
Post by: moekamo on June 02, 2011, 11:53:16 am: ahh, ok, in that case , heres what happened when i did it:

$\begin{align} \frac{dq}{dt}+10q & =\frac{1}{4}\sin5t \qquad I= e^{\int 10 dt} = e^{10t}\\ \implies e^{10t} \frac{dq}{dt} + 10e^{10t}q &= \frac{1}{4}\sin5t e^{10t}\\ \implies \frac{d}{dt}(qe^{10t}) & =\frac{1}{4} e^{10t}\sin5t \\ \implies qe^{10t} & = \int \frac{1}{4}e^{10t} \sin5t dt \qquad \mbox{Use product rule with } f=e^{10t}, g'=\sin 5t\\ &=-\frac{1}{20} e^{10t} \cos 5t + \frac{1}{2} \int e^{10t} \cos 5t dt \qquad \mbox{Use product rule with } u=e^{10t}, v'=\cos 5t\\ &= -\frac{1}{20}e^{10t} \cos 5t + \frac{1}{10}e^{10t} \sin 5t - \int e^{10t} \sin5t dt + C \\ &= -\frac{1}{20}e^{10t} \cos 5t + \frac{1}{10}e^{10t} \sin 5t -4qe^{10t} + C \\ \implies 5e^{10t}q&= -\frac{1}{20}e^{10t} \cos 5t + \frac{1}{10}e^{10t} \sin 5t +C \\ \implies q&= -\frac{1}{100} \cos 5t + \frac{1}{50} \sin 5t +Ce^{-10t} \end{align} $

which is the same as before, hope this helps find your mistake :)
Title: Re: integrating by parts
Post by: berryy on June 03, 2011, 04:41:12 pm: i know where i did wrong now
Thankyou heaps for the help :)