Laplace Transform?

Question

dy/dt = 4y + 9 t (e)^t
by laplace transform
i dont know how
9 t (e)^t >> 9 / (s-1)^2
....plz

Answer 1

Here are a few Laplace Transforms to know (use a table):

L.T. [dy/dt] ------> sY - y(0), where y(0) is the initial value
L.T. [9t*e^t] -----> = 9/(s-1)^2

Plugging into you differential equation, you get:

sY - y(0) = 4Y + 9/(s-1)^2, now solve for Y
Y(s-4) = 9/(s-1)^2 + y(0) ----->
Y = 9/[(s-1)^2(s-4)] + y(0)/(s-4)

To simplify the first term, you need to use partial fractions. This states that:

9/[(s-1)^2(s-4)] = A/(s-4) + B/(s-1) + C/(s-1)^2. This is the repeated root case. Multiply both sides by (s-1)^2(s-4). You get:

9 = A(s-1)^2 + B(s-1)(s-4) + C(s-4). Simplifying you get,

9 = A(s^2-2s+1) + B(s^2-5s+4) + C(s-4). Now group the variable and constant terms:

9 = s^2(A+B) + s(-2A-5B+C) + (A+4B-4C). Now compare coefficients and write equations for each:

s^2 terms:
0 = A+B (0 on left side because there are no s^2 terms)

s terms:
0 = -2A-5B+C(0 on left side because there are no s terms)

constant terms:
9 = A+4B-4C

You have 3 equations and 3 unknowns, now solve for A, B, C. Solving, you get:

A = 1, B = -1, C = -3 . Now plug into the above equation and you get:

9/[(s-1)^2(s-4)] = 1/(s-4) - 1/(s-1) - 3/(s-1)^2. Now use inverse Laplace to solve [remember that L.T. e^(at) = 1/(s-a)]

Using the original equation, you get:

y(t) = exp(4t) - exp(t) - 3t*exp(t) + y(0)*exp(4t)

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You also have a question about 9t*exp(t) ----> 9/(s-1)^2. You need to use the definition of Laplace Transform to solve:

int(0 -> inf): 9t*exp(t)*exp(-st)dt = 9t*exp[-t(s-1)]dt. You need to use integration by parts to solve:

int(u*dv) = uv - int(v*du)

u = 9t -----> du = 9*dt
dv = exp(-t[s-1]) ------> v = -1/(s-1)[exp(-t[s-1])]

Plugging above and evaluating you get:

9t/(s-1)[exp(-t[s-1])] + int(9/(s-1)[exp(-t[s-1])]). When evaluated from 0 to infinity, the first term goes to 0. Thus, the 2nd term becomes:

-9/(s-1)^2*exp[(-t[s-1])], evaluated from 0 to infinity. Doing that, you get:

9/(s-1)^2, which is the answer you are looking for.

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Hope this helps

Answer 2

Notation: The initial value is y(0). The Laplace transform of y is Y.

If you take the Laplace transform of both sides you get

L[y'] = L[4y] + L[9te^t]
sY-y(0) = 4Y + 9/(s+1)^2
(s-4)Y = y(0)+9/(s-1)^2
Y = y(0)/(s-4) + 9/((s-4)(s-1)^2)

The last term is expanded using partial fractions,
9/((s-4)(s-1)^2) = a/(s-1) + b/(s-1)^2 + c/(s-4)
a = -1, b=-3, and c=1

Y = y(0)/(s-4) -1/(s-1) - 3/(s-1)^2 +1/(s-4)

The inverse Laplace transform is
y = (y(0)+1)e^(4t) - e^t -3te^t

Answer 3

If this can help