Question

Solve the differential equation.

Answer 1

which one?

Answer 2

x''+ax'+bx=c cos(wt)

Answer 3

@experimentX you can read people's minds?

Answer 4

He is skilled in that domain, yes.

Answer 5

x''+ax'+bx = 0 <-- this part gives the complimentary function as usual.

Answer 6

there's no question

Answer 7

for particular integral ...
Write this as
\[ (D^2 + aD +b)x = c \cos ( \omega t)\]

case 1: you simply have no resonance.
\[ \omega^2 + a\omega + +b \neq 0\]
case 2: you have resonance
\[ \omega^2 + a\omega + +b = 0\]

let's do for case 1:

Answer 8

the auxiliary equation is
\[ r^ 2 +ar + b = 0 \implies r = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\]
case 1.1: \( \sqrt{a^2 - 4b } \in \Re\) this case it is simply overdamped

case 1.2 \( \sqrt{a^2 - 4b } \in \Im\) this case it is damped.

the complimentary functions are

1.1 \( \large x = c_1 e^{\frac{-a + \sqrt{a^2 - 4b}}{2}}+c_2 e^{\frac{-a - \sqrt{a^2 - 4b}}{2}}\)

1.2 \( x = e^{-a \over 2} (c_1 \cos \omega_0t + c_2 \sin \omega_0 t)\)

where \( \omega_0 = {\sqrt{ 4b - a^2}\over 2}\) is the frequency of system.

Answer 9

Woops ... worry here
\[ \large x = c_1 e^{\left( \frac{-a + \sqrt{a^2 - 4b} }{2}\right)t}+c_2 e^{\left(\frac{-a - \sqrt{a^2 - 4b}}{2}\right)t} \]

^^ Note the above case that \( a> \sqrt{a^2-4b} \) and this always decays with time t. this is characteristics of decay equation.
\[ \large x = e^{-a t\over 2} (c_1 \cos \omega_0t + c_2 \sin \omega_0 t) \]

^^ IN above case ... it harmonic but \( e^{-at/2}\) always damps it with passage of time.

Answer 10

For particular integral ... with the input type of Exponential
\[ \large f(D) x = e^{at}\]
\[ P.I. = \frac{e^{at}}{f(a)} \text{ if f(a) } \neq 0 \\
P.I. = \frac{t e^{at}}{f'(a)} \text{ if f'(a) } \neq 0 \\
..\\
P.I. = \frac{t^n e^{at}}{f^n(a)} \text{ if } f^n(a)\neq 0 \\
\]

Answer 11

In case 1.2, is c_2 be imaginary?

Answer 12

nope: they are just constants of integration.

Answer 13

if you have complex solution then ...
then the linear combination of both real and imaginary part (you have to exclude i) will give the general solution.

Answer 14

\[a^2-4b \le 0\]
\[x=c_1e^{-0.5at+0.5i\sqrt{4b-a^2}}+c_2e^{-0.5at-0.5i\sqrt{4b-a^2}}\]
\[x=e^{-0.5at}(c_1coswt+ic_1sinwt+c_2coswt-ic_2sinwt)\]
\[x=e^{-0.5at}(Acoswt+Bsinwt)\]

Answer 15

you can check that ...
\[ f(a + ib) = f(a) + if(b) = 0 \implies f(a) = 0, f(b) = 0 \text{ so both a and b are solution.} \]

Answer 16

\[* ((0.5i \pm \sqrt{4b-a^2})t)\]

Answer 17

\[ \large \cos ( \omega t) = \Re \left( e^{i \omega t}\right)\]
you can substitute it to the P.I. above and get it's particular integral.

Answer 18

Is my logic correct for why B should be complex?

Answer 19

okay .. B is complex. do you know what is a linear map?

Answer 20

or linear function?

Answer 21

Yes, I understood your reasons why f(a)=0 and f(b)=0 are both solutions. But as this is to do with an oscillator, I guess you would disregard the complex part.

Answer 22

But I don't understand how you arrived at the earlier form for the PI

Answer 23

In this specific problem, for instance, how would you calculate the PI?

Answer 24

both \( x_1 \) and \( x_2 \) are solution to your system ... from superposition principle ... any linear combination of \( x_1 \) and \( x_2 \) should be solution to your system. so B doesn't necessarily have to be complex.

Answer 25

driven damped harmonic oscillator :)

Answer 26

for this particular problem ... it's quite easy.
\[ P.I. = \frac{e^{at}}{f(a)} \text{ if f(a) } \neq 0 \\
P.I. = \frac{t e^{at}}{f'(a)} \text{ if f'(a) } \neq 0 \\
..\\
P.I. = \frac{t^n e^{at}}{f^n(a)} \text{ if } f^n(a)\neq 0 \\ \]

and
\[ \large \cos ( \omega t) = \Re \left( e^{i \omega t}\right) \]

Answer 27

and \[ f(D) = D^ 2 + aD + b\]
Woops!! should have put 'p' instead of 'a' since we have 'a' already.

Answer 28

So to calculate the PI, we need to find an f(a) such that:\[\frac{f'(a)}{f(a)}=t\]?

Answer 29

\[
P.I. = \Re \left(\frac{c e^{i \omega t}}{f(i \omega)} \right ) \text{ if } f(i\omega) \neq 0 \\
P.I. = \Re \left(\frac{c t e^{i \omega t}}{f'(i \omega)} \right ) \text{ if }f'(i \omega) \neq 0 \\
P.I. = \Re \left(\frac{c t^2 e^{i \omega t}}{f''(i \omega)} \right ) \text{ if } f''(i\omega)\neq 0 \\
\]

Answer 30

lol ... I didn't realize we would have 3 cases of PI

Answer 31

What is f(iw) in this case?

Answer 32

f(D) = D^2 + aD + b
you put iw inplace of D and check if it's zero or not.

Answer 33

suppose this is not zero ... then your particular integral is simply
\[ P.I. = \Re \left(\frac{c e^{i \omega t}}{(i \omega)^2 + a (i \omega) + b} \right )\]

Answer 34

this is not resonant case.
if \( (i \omega)^2 + a (i \omega) + b = 0\), then you have resonance so your PI would be
\[ P.I. = \Re \left(\frac{tc e^{i \omega t}}{2(i \omega) + a } \right ) \]
this is destructive http://www.wolframalpha.com/input/?i=plot+t+sin%28t%29+from+0+to+8+pi

Answer 35

you can solve for PI with method of undetermined coefficients or reduction of order or variation of parameters
the approach i used above is quite unintuitive you can find the proof here

it's been discussed here in detail
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/exponential-response/

the general proof is on this document.

http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf
from this chapter http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/

Also

Answer 36

try using try using
http://en.wikipedia.org/wiki/Variation_of_parameters
http://en.wikipedia.org/wiki/Reduction_of_order
http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients
they are pretty straightforward and intuitive.

Answer 37

I'll have a look at those, but keep the question open just in case- Thank you

Answer 38

sure ... also try to put the equation on your first post.