Question

Find the solution of y'' + 4y = 3t, y(0) = 0, y'(0) = 0

Answer 1

Use variation of parameters or undefined coef method?

Answer 2

\[y'' + 4y = 3t\]\[r^2 + 4 = 0\]\[r=\pm 2i\]\[0=\alpha, 2=\beta\]This is where I am

Answer 3

I'm going to post my Yh if somebody could check it

Answer 4

\[y_h=c_1e^{(0)t}\sin2t+c_2e^{(0)t}\cos 2t\]

Answer 5

\[y_h=c_1\sin2t+c_2\cos 2t\]

Answer 6

i dont think yh would be this.. i think sin and cos are in roots when complex roots.

Answer 7

I have complex roots

Answer 8

oops sorry ddnt saw it.

Answer 9

Try using variation of parameters where Y1 = sin(2t) and Y2 = cos(2t). and use the wronskian. Then plug it into

\[-Y _{1}\int\limits(\frac{ y _{2}g(t) }{ W })dt+Y_{2}\int\limits \frac{ y_{1}g(t) }{ W }dt\]

Answer 10

where "W" is wronskian.

Answer 11

honestly I haven't seen that. Can we stick with this method? I think my next step is to solve for c1 and c2 given my initial conditions

Answer 12

No, before you can solve for c1 and c2, we need the particular solution.

Answer 13

Oh ya, Yp

Answer 14

Oh wait, zepdrix is right. I didn't see that it's complex. Sorry.

Answer 15

try going with undetermined cofficient .. i think would be easy.

Answer 16

I need to look through my notes a sec. I don't remember how to find Yp

Answer 17

multiplicity (s) = 0 correct?

Answer 18

yes. use undet coef and make a guess for 3t to solve and get ur particular solution

Answer 19

Use Ct as your guess. Take y' and y'' and plug it into your solution and set it equal to Ct. Solve for C.

Answer 20

Your particular solution will be of a form that matches the right side of your equation. In this case since we have a polynomial, our Yp will be a polynomial starting at degree 1,\[\huge y_p=At+B\]

Answer 21

\[ay"+b'+cy=Ct^me^{rt}\]\[y_p=t^se^{rt}(At^m+...+A_0t^0)\]I got that too zep. \[y_p=At+B\]

Answer 22

so my next step is to use Yp and the initial conditions to find A and B

Answer 23

Mmmmm you should be able to find A and B without the initial conditions :D hmm

Answer 24

\[y_p=At+B\]\[y_p'=A\]I'm stuck A=0 B=0 that doesn't seem right

Answer 25

\[\large y_p=At+B, \quad y_p^{''}=0\]Plugging these into our equation gives us,\[\large y^{''}+4y=3t \quad \rightarrow \quad 0+4(At+B)=3t\]

Answer 26

Is my Yp wrong because it is complex

Answer 27

That is your Yc :) the complimentary solution. We don't want to mix that in with our Yp until we've found A and B.

Answer 28

:o

Answer 29

Ok I follow you @zepdrix

Answer 30

\[\large y_c=c_1 \sin 2t+c_2 \cos 2t\]Yp will be of a form that matches the right hand side :D
\[\large y_p=At+B\]

Good now? k :3

Answer 31

Just to remind you what we're trying to get :D\[\large y=y_c+y_p\]

Answer 32

\[At+B=\frac{ 3 }{ 4 }t\]Now use my initial conditions

Answer 33

B=0 A=3/4 ?

Answer 34

Yah looks good! :)

Answer 35

And NOWWWWW, after you get it into this form,\[\large y=y_c+y_p\]You can FINALLY use your initial conditions \:D/

Answer 36

I want to go back to solving for A and B

Answer 37

When I put Yp's into the original how do I solve for A and B. By equating terms?

Answer 38

Yesss, remember how you do that in Partial Fractions? :) It's prretty much the same process.

Answer 39

Ok that makes sense. I got lucky earlier :))

Answer 40

Oh lol :D

Answer 41

\[y=c_1 \sin 2t+c_2 \cos 2t+\frac{ 3 }{ 4 }t\]

Answer 42

smooth.

Answer 43

\[0=c_1 \sin 2(0)+c_2 \cos (0)t+\frac{ 3 }{ 4 }0\]\[0=c_2\]\[y'=2c_1 \cos2t+{3 \over 4}\]\[0=2c_1+{3 \over 4}\]\[c_1=-\frac{ 3 }{ 8 }\]\[y=-\frac{ 3 }{ 8 }\sin2t+\frac{ 3 }{ 4 }t\]^^Final Answer^^

Answer 44

Yay good job \:D/

Answer 45

Thanks man. I have my Diff EQ final Wed. When is yours? If I remember you're in this class right now too?

Answer 46

Yah my Final is actually Wednesday also XD

I have a pretty good teacher for this though, he keeps things rather simple :| it's almost been too easy at times.

Spending most of my time preparing for Physics Final Exam tomorrow, and Calc 3 on Friday D: I'm not even worried about the DIff EQ Final.

Answer 47

You shouldn't be. Strange how you're in Calc 3 at the same time as Diff EQ.

My teacher is really smart but he just stares at the board and does problems. He does like 3 steps in his head at once sometimes and we all just look at eachother like WTF just happened.

Good luck on your Physics final. I like physics