Question

Find the IVP by using Laplace Transforms y''+5y'-6y=21e^{t-1}; y(1)=-1, y'(1)=9

Answer 1

well, take the laplace ... what do we get?

Answer 2

Well i'm not sure if i need a translation on the t-axis? Taking the Laplace of both sides is fairly easy, but given the fact that the initial conditions are not at zero, i'm not sure what to do at the moment.

Answer 3

yeah, thats a rare twist ... let r = t-1 if you want to substitute it

Answer 4

im not that familiar with what a convolutionis so im not sure if that would be useful here

Answer 5

y(t-1) = y(r) .... y(t=1) = y(r=0) is what im thinking

Answer 6

If L{f(t)} exists for s>c, then
L{u(t-a)f(t-a)}= e^{-as}F(s) for s>c+a

Answer 7

I think this may be it

Answer 8

by definition
\[L\{y''(t-1)\}=\int_{0}^{\infty}y''(t-1)e^{-st}~dt \]
let t = r+1, dt = dr
\[L\{y''(r+1-1)\}=\int_{0}^{\infty}y''(r+1-1)e^{-s(r+1)}~dr \]
\[L\{y''(r)\}=e\int_{0}^{\infty}y''(r)e^{-sr}~dr \]

Answer 9

your find may be it, im just not that adept at all the transforms since i dont get to use it that often :)

Answer 10

ive made an error in my thought ... e^(-sr) e^(-s)

so e^(-s) pulls out as a constant

Answer 11

your method looks adequate; you're using convolution. I may try that method.

Answer 12

since i need the practice :) which means this could get messed up lol

\[L(y'')+5L(y')-6L(y)=21L(e^{r})\]
\[L(y'')+5L(y')-6L(y)=21L(e^{r})\]
\[s^2L(y)-sy(0)-y'(0)+5(sL(y)-y(0))-6L(y)=21L(e^{r})\]
\[s^2L(y)-sy(0)-y'(0)+5sL(y)-5y(0)-6L(y)=21L(e^{r})\]
\[s^2L(y)+s-9+5sL(y)+5-6L(y)=21L(e^{r})\]
\[L(y)(s^2+5s-6)=21L(e^{r})+4-s\]
\[L(y)=\frac{25-s}{(s+6)(s-1)^2}\]

Answer 13

21/(s-1) +4 - s so i messed up the last 2 rows i believe but its fixable lol

Answer 14

17 +5s -s^2 might be the numerator ... if you want to let me know when and where i mess this totally up feel free :)

Answer 15

Sounds good, ill let you know my answer in a bit.

Answer 16

\[17 +5s -s^2=A(s-1)^2+(Bs+C)(s+6)\]
\[17 +5(-6) -36=A(-6-1)^2\]
\[-49=49A~:~A=1\]
\[17 +5s -s^2=(s-1)^2+(Bs+C)(s+6)\]
\[17 +5 -1=(B+C)(1+6)\]
\[3=B+C\]
\[17=(0-1)^2+C(0+6)\]
\[8/3=C~:~B=1/3\]
\[L(y)=\frac{1}{s+6}+\frac13\frac{s}{(s-1)^2}+\frac83\frac{1}{(s-1)^2}\]
\[y(r)=e^{-6r}+\frac13 e^r(r+1)+\frac83e^rr\]
\[y(t-1)=e^{-6(t-1)}+\frac13 e^{t-1}t+\frac83e^{t-1}(t-1)\]
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the wolf gives us a solution as:
\[y(t) = e^{-6t-1} (3 e^{7t} (t-1)-e^7)\]
\[y(t) = 3e^{6t-1} (t-1)-e^{-6t+6}\]
so im pretty sure ive messed something up along the way, of course the wolf solved for y(t) which would be a shift of our y(t-1) maybe?

Answer 17

wow! This one is brutal.

Answer 18

yeah, and i think my decomp is messed up ...i hate typing and trying to math

Answer 19

A = -1 not 1 messed up the rest of the process

Answer 20

C=3

Answer 21

B=0

Answer 22

\[L(y)=3\frac{1}{(s-1)^2}-\frac{1}{s+6}\]
\[y(r)=3r~e^r-e^{-6r}\]
i wonder if i should put back on the e^-s, when i do the wolf gives me some heaviside function

Answer 23

the wolf agrees with \[y(r) = 3r e^r-e^{-6r}\] :)

Answer 24

I think they are using Duhamel's Principle. They're solving for the general solution then

Answer 25

applying convolution

Answer 26

good job.

Answer 27

And thanks for the help.

Answer 28

good luck

Answer 29

Likewise amistre64. It was nice talking to you. Cheerio!