Question

Use laplace transform to solve the differential equation:

d^2y/dt^2-4dy/dt+4y=7e^-2t*sin3t-24e^-2t*co3t

conditions
y(0)=0 and dy/dt=3 t =0

Answer 1

Hi i have attempted it up to:

s^2L(X)-3-4SL(X)+4Y=7/S+2*-3/S^2+3^2-24/S+2*2/S^2+3^2

i DONT KNOW HOW TO PROCEED FURTHER?

Answer 2

I transformed the RHS first... so it becomes complicated!

Answer 3

Any ideas?

Answer 4

sry dont know how to solve this...what grade r u in?

Answer 5

University engineering...

Answer 6

lol why did you view the question then ? ?

Answer 7

i was trying to understand it ....duh.

Answer 8

\[\large \mathcal L \left\{\frac{ d^2y }{ dt^2 }-4\frac{ dy }{ dt }+4y\right\}=\mathcal L \left\{7e^{-2t}\sin3t-24e^{-2t}\cos 3t\right\}\]

Answer 9

s^2L(X)-3-4SL(X)+4Y=7/S+2*-3/S^2+3^2-24/S+2*2/S^2+3^2

Answer 10

yes sirm thats the correct questions....

Answer 11

There you go.....youve got someone to help you! Gud luck^^

Answer 12

\[\large s^2\mathcal L\left\{y\right\}-3-4\mathcal L \left\{y\right\}=7\left( \frac{ 3 }{ (s+2)^2+9 } \right)-247\left( \frac{ s+2 }{ (s+2)^2+9 } \right)\]

Answer 13

as you see i took the laplace transform on the rhs to make it less diffcult... is my first line of working correct.?

Answer 14

\[\mathcal L \left\{ e^{at}f(t)\right\}=F(s-a)\]

Answer 15

sirm where did you get 247 from?

Answer 16

7 is the coefficient of s&{-2t}sint 3t, 7 is for the term with cosine