Question

can anyone help me solve this initial value problem http://dl.dropbox.com/u/530879/snip.PNG ?

Answer 1

ima try laplace, im sure Noliec is gonna go traditional tho :)

Answer 2

I'm not 100% sure as to what I'm doing so I do not dare to post, after Mr legendary amistre64 came along! :D

Answer 3

i dont think thats how it works :)

Answer 4

Eagerly awaiting your explaination amistre!

Answer 5

y''-y'+2y=4cos(2t) + 2sin(2t)

solve the homogenous with the characteristic eq;

r^2-r+2=0

r^2-r=-2

r^2-r+1/4=-2+1/4

(r-1/2)^2= -7/4; r=(1+- i sqt(7))/2

yh = \(e^x(c_1cos(\sqrt{7}x)+c_2sin(\sqrt{7}x))\)

Answer 6

y = yh + yp ; since yh=0, we find yp

Answer 7

Ah! About that, how are you supposed to remember all the different forms for differential equations?

Answer 8

You got any aide-memoire or just genius? :D

Answer 9

dunno, i havent come across them all :)

Answer 10

they are all pretty much the same tho

Answer 11

there are 2 ways that i know of to get the yp; a long way and a short way .. both are rather long tho

Answer 12

ima try the wronskian ...

\[y_p=y_1\int \frac{W_1}{W}+y_2\int \frac{W_2}{W}\]

Answer 13

\[\begin{vmatrix}
W_1&e^xcos(\sqrt{7}x)&e^x(cos(\sqrt{7}x)-\sqrt{7}sin(\sqrt{7}x)\\
W&0& 4cos(2x)+2sin(2x) & \\
W_2&e^xsin(\sqrt{7}x)&e^x(sin(\sqrt{7}x)+\sqrt{7}cos(2x))
\end{vmatrix}\]

that fits, so now we do a psuedocross

Answer 14

yp being the particular yes

Answer 15

\[L\{y''-y'+2y=4cos(2t)+2sin(2t)\}\]

\[s^2Fs-sy''(0)-y'(0)-sFs+y'(0)+2Fs=\frac{4s+2(2)}{s^2+4}\]

\[s^2Fs-s-1-sFs+1+2Fs=\frac{4s+4}{s^2+4}\]

\[Fs(s^2-s+2)=\frac{4s+4}{s^2+4}+s\]

\[Fs=\frac{\frac{4s+4}{s^2+4}+s}{s^2-s+2}\]

Answer 16

\[Fs=\frac{s^3+8s+4}{(s^2+4)(s^2-s+2)}\]

we do partials to that and work the inverse laplace

this would be an alternative way

Answer 17

http://www.wolframalpha.com/input/?i=y%27%27-y%27%2B2y%3D4cos%282x%29%2B2sin%282x%29

it loks like i missed the yh by a 1/2

Answer 18

oh i got it, i just forgot to include it on the next one lol

Answer 19

http://www.wolframalpha.com/input/?i=y%27%27-y%27%2B2y%3D4cos%282x%29%2B2sin%282x%29%2C+y%27%280%29%3D1%2C+y%27%27%280%29%3D1

it looks like its gonna be a pain no matter what; so go thru the steps that they offer and if you have any specific questions, feel free to ask.

Answer 20

You deserve 10 medals!