Question

y''-7y'+10y=9cost+7sint y(0)=5, y'(0)=-4
Laplace transform IVP
ive got it to the point of partial fraction decomp but know im gonna make a mistake here please ensure that i dont

Answer 1

you can do it! :D

Answer 2

lol

Answer 3

\[Y(s)(s^2-7s+10)=s^2/(s^2+1)+7/s^2+1+5s-39\]

Answer 4

i know that
\[s^2-7s+10\]
becomes
\[(s-5)(s-2)\]
but its the right side i know ill screw up

Answer 5

i was thinking maybe its
\[(s^2+5s-32)(s^2+1)\]
but not sure

Answer 6

oh all over s^2+1

Answer 7

@Psymon from the problem earlier with the w substitution we had to do w-1 because they gave our IVP w(-1)=4 when we have IVP's with laplace transforms the IVP have to be 0 like y(0)=3, y'(0)=7

Answer 8

completely forgot about that

Answer 9

Yeah, I recall reading that. Didnt remember what the problem exactly was, but yeah, something unique to be aware of.

Answer 10

hmm:
$$
y''-7y'+10y=9\cos t+7\sin t\qquad y(0)=5, y'(0)=-4\\s^2Y-sy(0)-y'(0)-7(sY-y(0))+10Y=\frac{9s}{s^2+1}+\frac7{s^2+1}\\s^2Y-7sY+10Y-5s+4+35=\frac{9s+7}{s^2+1}\\(s^2-7s+10)Y(s)=\frac{9s+7}{s^2+1}+5s-39\\(s^2-7s+10)Y(s)=\frac{9s+7+5s(s^2+1)-39(s^2+1)}{s^2+1}\\(s^2-7s+10)Y(s)=\frac{9s+7+5s^3+1+5s-39s^2-39}{s^2+1}\\(s-5)(s-2)Y(s)=\frac{5s^3-39s^2+14s-32}{s^2+1}\\Y(s)=\frac{5s^3-39s^2+14s-32}{(s-5)(s-2)(s^2+1)}$$

Answer 11

yes @rperez36 because the Laplace transform considers information only from \(0\) to \(\infty\) all IVPs need to be 'time shifted' so that the initial values correspond to \(t=0\)

Answer 12

oops i have a stray \(1\) on my 6th line

Answer 13

ah ok so i did need to keep the 9. @oldrin.bataku thanks again i cant thank you enough you will be the reason i pass this class. test on thur and trying to get ahead of the class. @Psymon give the man his medal. Its the least we can do i feel like i should be paying you for this wisdom you bestow upon me lol

Answer 14

Yeah, when he's answering a question I kinda just shut up and wait for his wisdom, haha.

Answer 15

no doubt. now lets see if i can finish this beast

Answer 16

I should catch up to ya in all the D.E topics, I've fallen behind being distracted by other stuff, lol.

Answer 17

$$Y(s)=\frac{5s^3-39s^2+14s-32}{(s-5)(s-2)(s^2+1)}=\frac{A}{s-5}+\frac{B}{s-2}+\frac{Cs+D}{s^2+1}\\A(s-2)(s^2+1)+B(s-5)(s^2+1)+(Cs+D)(s-2)(s-5)\\\ =A(s^3-2s^2+s-2)+B(s^3-5s^2+s-5)+(Cs+D)(s^2-7s+10)\\\ =(A+B)s^3+(-2A-5B)s^2+(A+B)s+(-2A-5B)\\\ \ \ \ \ \ \ \ \ \ +(Cs^3+(-7C+D)s^2+(10C-7D)s+10D)\\\ =(A+B+C)s^3-(2A+5B+7C-D)s^2+(A+B+10C-7D)s\\\ \ \ \ \ \ \ \ \ \ -(2A+5B+10D)$$hence we get:$$A+B+C=5\\2A+5B+7C-D=39\\A+B+10C-7D=14\\2A+5B+10D=32$$taking the difference between the second and fourth we get:$$7C-11D=7$$and the first and third gets us:$$9C-7D=9$$eliminating between these two yields \(C=1,D=0\) and from this we find:$$A+B=4\\2A+5B=32$$hence \(A=-4,B=8\) and ultimately we conclude:$$Y(s)=-\frac4{s-5}+\frac8{s-2}+\frac{s}{s^2+1}$$

Answer 18

to finish it off we just take the inverse transform:$$y(t)=\mathcal{L}\left\{-\frac4{s-5}+\frac8{s-2}+\frac{s}{s^2+1}\right\}=-4e^{5t}+8e^{2t}+\sin t$$

Answer 19

oh wait that should be \(\cos t\) oopsies

Answer 20

thank you! working it i was able to get C and D but but i messed up my signs so A and B i couldn't get i just think its bed time. Thanks a lot @oldrin.bataku

Answer 21

no problem! glad you figured it out :-p