Question

Use Laplace transformations to solve the IVP:
x''+6x'+9x=7e^(4t) x(0)=0, x'(0)=0

Answer 1

well, we simply need to get our hands dirty and take the Laplace to start with

Answer 2

ideally, we should have built up a table of laplace transforms to be able to speed up the process.

Answer 3

yes agree with you

Answer 4

\[L[\frac{ dy }{ dt }]=sY(s)-Y(0)\]

Answer 5

\[\int_{0}^{inf} fe^{-st}~dt\]

making a table of by parts

e^(-st)
+ f -e^(-st)/s
- f' e^(-st)/s^2
+ f'' -e^(-st)/s^3
- f''' e^(-st)/s^4
....

we can develop the general setup

Answer 6

After doing the transformation, this is what i got and am not sure about it
\[y=\frac{ s^2-4s+7 }{ (s-4)(s+3)^2 }\]

Answer 7

\[L[Y'']=s^2Y(s)-sY(0)-Y'(0)\]

Answer 8

now lets apply LT to the equation

Answer 9

yes, the inverse :)

Answer 10

\[x''+6x'+9x=7e^{4t}\\apply~LT\\s^2X(s)-sX(s)-X'(0)+6sX(s)-X(0)+9X(s)=\frac{ 7 }{ s-4 }\]

Answer 11

seems ok?

Answer 12

now put the initial conditions

Answer 13

does that look okay
\[(s^2+6s+9)X=\frac{ 7 }{ s-4 }\]

Answer 14

\[X=\frac{ 7 }{ ((s-4)(s+3)^2 }\]

Answer 15

\[s^2X(s)-sX(s)+6sX(s)+9X(s)=\frac{ 7 }{ s-4 }\\X(s)(s^2+5s+9)=\frac{ 7 }{ s-4 }\]

Answer 16

u missed a s it should be 5s not 6s okay?

Answer 17

Doesnt that s cancel since we have x(0)=0 for the initial value.

Answer 18

oh yes sorry my bad ,

Answer 19

so u have now\[X(t)=L^{-1}\frac{ 7 }{( s-4)(s+3)^2 }\]
now we have to use partial fraction

Answer 20

Then how do i get A, B and C
\[\frac{ A}{ s-4 }+\frac{ B }{ s+3}+\frac{ C }{ (s+3)^2 }\]

Answer 21

or we can use convolution ,are u familiar with it?

Answer 22

I havent heard of that.

Answer 23

ok lets do partial fraction ,ure attemp is correct

Answer 24

ok can u do it ? the expansion?

Answer 25

So far av been able to get A=1. The rest am confused how to go about it

Answer 26

S

Answer 27

\[\frac{ 7 }{ (s-4)(s+3)^2}=\frac{ A}{ s-4 } +\frac{ B }{ s+3 }+\frac{ c }{( s+3)^2 }\\ \frac{ A.(s+3)^2+B.(s-4)(s+3)+C(s-4) }{ (s+3)^2.(s-4) }=\frac{ 7 }{ (s-4)(s+3)^2 }\\A(s+3)^2+b(s+3)(s-4)+C(s-4)=7\\now ~put~,s=-3\\we~get\\C(-3-4)=7\\C=-1\]

Answer 28

now wait a bit for A,B

Answer 29

now \[put~s=4~we'\ll ~have\\A(3+4)^2=7\\A=1/7\]

Answer 30

now forB\[put~s=1\\A(1+3)^2+B(1+3)(1-4)+C(1-4)=7\\ \frac{ 16 }{ 7 }-12B+3=7\\B=\frac{ -1 }{ 7 }\]

Answer 31

\[so~We~have\\A=1/7\\B=-1/7\\C=-1\]

Answer 32

ok????

Answer 33

now i believe u can do the rest

Answer 34

So far it looks okay. I had made an algebra error on A but i got it now

Answer 35

ok u can take it from here?

Answer 36

Yes and now i can be able to do the rest of the homework. Thanks alot

Answer 37

no problem buddy :)