laplace transformation question?

Question

find the solution y(t) for
y'' - 2y' + 7y = sin(t) y(0)= 0 y'(0) = 0

I was able to reach the laplace transform of y(t) as:
1/ (s^2 + 1) ( s^2 - 2s + 7)

but i have no idea how to invert it back into y(t), please help me

Answer 1

Yes, partial fractions are your friend here:
Y(s) = 1/ [(s^2+1) (s^2-2s+7)]
= (As + B) / (s^2 + 1) + (Cs + D) / (s^2 - 2s + 7)
<=> (As + B) (s^2 - 2s + 7) + (Cs + D) (s^2 + 1) = 1
<=> (A+C) s^3 + (B-2A+D) s^2 + (7A-2B+C) s + (7B+D) = 1
so A+C = 0, B-2A+D = 0, 7A-2B+C = 0, 7B+D = 1
which gives
C = -A, B + D = 2A, B = 3A, 7B + D = 1
and hence C = D = -A, B = 3A, 20A = 1
whence
Y(s) = 1/20 [(s + 3) / (s^2 + 1) - (s + 1) / (s^2 - 2s + 7)]
Noting that s^2 - 2s + 7 = (s-1)^2 + 6, we get
y(t) = 1/20 [cos t + 3 sin t - e^t cos (√6 t) - 2/√6 e^t sin (√6 t)]

You can verify that this is the correct answer.

Answer 2

The standard method is to use a PFE (Partial Fraction Expansion) to get it into something tractable, i.e. a sum of low-order polynomials whose inverse-transforms are known.

The general inverse-transform you want here is
(s-α)/((s-α)² + ω²) → e^αt cos(ωt) U(t-α)

1/(s² + 1) is the simple case α=0, ω=1 → sin (ωt) U(t) with w=1

[When α>0, we are using the Shifting Theorem on this basic form F(s-α) → e^αt f(t) U(s-α)]
For 1/(s² - 2s + 7), we rewrite it as 1/((s-1)² + 6)

So the PFE is:
1/(s² + 1)( s² - 2s + 7) = (As+B)/(s² + 1) + (Cs+D)/( s² - 2s + 7)
Multiply across by the denominator:
1 = (As+B)(s² - 2s + 7) + (Cs+D)(s² + 1)
Before we go further, note there is no s³ or s² term, => C= -A
(As+B)(s² - 2s + 7) + (-As+D)(s² + 1) =1
As³ +(B-2A)s² +(7A-2B)s + 7B -As³+Ds²-As+D = 1
(B-2A+D)s² +(6A-2B)s + (7B+D) = 1
Casting out coefficients in [s^n terms]:
[s²]: (B-2A+D)=0
[s]: (6A-2B)=0 => A=B/3
[1]: (7B+D)=1 => D=1-7B

B-2B/3 +1-7B=0
20B/3 = 1
B = 3/20
A = 1/20
C = -1/20
D = -1/20

So we have:
1/(s² + 1)(s² - 2s + 7)
= 1/20 [ (s+3)/(s² + 1) - (s+1)/(s² - 2s + 7) ]
= 1/20 [ (s+3)/(s² + 1) - (s+1)/((s-1)² + (√6)²) ]
= 1/20 [ s/(s² + 1) +3/(s² + 1) - (s-1)/((s-1)² + (√6)²) + 2/((s-1)² + (√6)²) ]

Hence L^-1 [ 1/(s² + 1) /(s² - 2s + 7) ]
= 1/20 [ cos(t) U(t) + 3 sin(t) U(t) - e^t cos(t) U(t-1) + 2/√6 e^t sin(t) U(t-1) ]
= 1/20 [ [cos(t) + 3 sin(t)] U(t) + e^t [ -cos(t) + 2/√6 sin(t) ] U(t-1) ]

Answer 3

because of the fact that g(t) = 0 whilst a million <= t , the Laplace vital: int_{0}^{infty} ( exp( -s*t ) * g(t) ) dt = int_{0}^{a million} ( exp( -s*t ) * (t^3) ) dt Use integration via areas thrice to kill the t^3 element of the integrand. on the different hand, you ought to apply the step function U(t) and say that g(t) = ( t^3 ) * [ U(t) - U(t-a million) ] the place U(t) = 0 whilst t<0 and U(t) = a million whilst t>=0. desire that helps.

Answer 4

u sohould u se partial fractions for this.

And as soon as u got partial fractions,apply inverse laplase transform,then u get ur answer.
for examples of partial fractions ill giv u a link
(ignore the integration symbol.)
http://archives.math.utk.edu/visual.calculus/4/partial_fractions.2/

Answer 5

try the following link

http://rbmix.com/problem/ilpla/il.php