Question

Sole the differential equation

Answer 1

\[y''+9y=cos(3t)\]
\[y(0)=2\]
\[y'(0)=5\]

\[\mathcal L \{y'' \} + 9\mathcal L \{y \}=\mathcal L \{cos(3t)\}\]

Answer 2

\[s^2Y(s)-sf(0)-f'(0)+9Y(s)=\frac{s}{s^2+9}\]

Answer 3

\[s^2Y(s)-2s-5+9Y(s)=\frac{s}{s^2+9}\]

Answer 4

\[s^2Y(s)+9Y(s)=\frac{s}{s^2+9}+2s+5\]

Answer 5

\[Y(s)[s^2+9]=\frac{s}{s^2+9}+2s+5\]

Answer 6

\[Y(s)=\frac{s}{(s^2+9)^2}+\frac{2x}{s^2+9}+\frac{5}{s^2+9}\]

Answer 7

\[\mathcal L^{-1}\{Y(s)\}=\mathcal L^{-1} \{\frac{s}{(s^2+9)^2}\}+\mathcal L^{-1}\{\frac{2s}{s^2+9}\}+\mathcal L \{ \frac{5}{s^2+9}\}\]

Answer 8

\[y(t)=\mathcal L^{-1}\frac{s}{(s^2+9)^2}\} +2cos(3t)+\frac{5}{3}sin(3t)\]

Answer 9

\[\mathcal L^{-1}\{\frac{s}{(s^2+9)^2}\}=\mathcal L^{-1}\{\frac{s}{s^2+9}\frac{1}{s^2+9}\}\]

Answer 10

\[f(t)=cos(3t)\]
\[g(t)=\frac{1}{3}sin(3t)\]

Answer 11

\[\mathcal L^{-1}\{\frac{s}{s^2+9}\frac{1}{s^2+9}\}=\frac{1}{3}\int_0 ^t cos(3\tau)sin(3\tau)d\tau\]

Answer 12

@waterineyes

Answer 13

\[\frac{1}{3}\int_0^t cos(3\tau)sin(3(t-\tau))d\tau\]

Answer 14

error in my last line

Answer 15

see any other errors?

Answer 16

i need an trig identity for htis... .can't rememberi t though lol

Answer 17

Actually I am not very good at Laplace also I know some of its concepts.


I checked all you are doing right but I am not sure about how to solve that last part.

If you are doing that right then, I see what I can do for you in that part..

Answer 18

the first two parts are correct

Answer 19

not sure about the last laplace that i'm doing

Answer 20

Yeah I too not sure about that..

Answer 21

@eliassaab @jim_thompson5910

Answer 22

Do you have to use LaPlace?

Answer 23

I believe superposition would be much easier here.

Answer 24

you have to use laplace

Answer 25

oh never mind my suggestion then, pardon me but I am too tired for another Laplace Transform *smirks* (-: I will read through it once though.

Answer 26

my book has this example below but it doesn't explain how it comes to it -.-

Answer 27

I got also one formula but it will take some time to explain that or even this is possible that I can't.
Ha ha ha..

Answer 28

they end up with two sin functions like

\[sin(k\tau)sink(t-\tau)\]

and use the

sinAcosB formula... but i don't understand how is sink(t-tau) is cos(b)?

Answer 29

\[\mathcal{L}(\frac{s}{(s^2 + a^2)^2}) = \frac{1}{2a} t \cdot \sin(at)\]

Answer 30

Sorry \(\mathcal{L^{-1}}\)..

Answer 31

I think I can explain it..