Question

Laplace Transformation to solve Initial value problem

Answer 1

\[y \prime \prime +4y \prime= Cos(t-3)+4t y(3)=0, y \prime(3)=7\]

Answer 2

@lgbasallote @FoolForMath @satellite73 @experimentX @

Answer 3

Plz help me figure out this one

Answer 4

@phi @dpaInc

Answer 5

@nbouscal

Answer 6

@experimentX you can try

Answer 7

shouldn't there be two initial values??
f(0) and f'(0)

Answer 8

lol that was the main thing... that we have to solve for f(3)

Answer 9

Oh ,,,

Answer 10

didn't see ... haven't done it though i'll try

Answer 11

well people were saying replace t with t+3

Answer 12

@eigenschmeigen try?

Answer 13

Umm ... how is it gong to look like??
\[ \int_0^\infty f''(t+3) e^{-st} ds = \]

Answer 14

well try this way i have no freakin idea

Answer 15

@eliassaab may be able to help with this.

I'd have to learn a few things that I don't know yet, personally.

Answer 16

whoa prof is here

Answer 17

\[ \left [ e^{-st}\int f''(t+3) dt\right ]_0^\infty - \int_0^\infty y'(t+3)e^{-st} dt \]

Answer 18

hey i have resolved it

Answer 19

lol ... did you get it??

Answer 20

yeh

Answer 21

Oh .. great!!

Answer 22

Your final answer should look like
\[
y(t)=\frac{1}{272} e^{-4 t} ( 136 e^{4 t}
t^2-68 e^{4 t} t-731 e^{4 t}- \\ 16 e^{4 t} \sin
(3) \sin (t)-16 e^{4 t} \cos (3) \cos (t)+\\64
e^{4 t} \cos (3) \sin (t)- 64 e^{4 t} \sin (3)
\cos (t)-289 e^{12}+16 e^{12} \sin ^2(3)+16
e^{12} \cos ^2(3) )
\]

Answer 23

In simplified form
\[
y(t)=\frac{1}{272} \left(136 t^2-68 t-273 e^{12-4
t}-64 \sin (3-t)-16 \cos (3-t)-731\right)
\]