given
y(t)= e3t+cos6t-e3tcos6t
Now by using the standard laplace transfrom relations
L(e^(at)) = 1/(s-a)
L(cos(at))= s/(s2+a2)
L(eatcos(bt))= (s-a)/((s-a)2+b2)
Now by applying the above relations to the present question, We get
L(y(t))=Y(s) = [1/(s-3)]+[s/(s2+36)] - [(s-3)/((s-3)2+36)]
so comparing the above equation with the equation given in the question we get
Unknown Variable | Value |
a | 1 |
b | -3 |
h | 1 |
k | 36 |
p | *** |
q | -6 |
t | 45 |
***There is a small issue in finding value of 'p' because it cannot be found in the given form of the expression.
In the following image the correct explaination with steps is given.
NOTE:
Please do comment to get clarified in any part of the solution.
The value of 'p' cannot be found out with the given expression.
There should be a -ve sign infront of the 3rd term, If it is the case then the value of 'p' is -3.
Thank you.
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for
part A which answer is correct?
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Consider the following IVP
y″ + 5y′ +
y = f (t), y(0) = 3,
y′(0) = 0,
where
f (t) =
{
8
0 ≤ t ≤ 2π
cos(7t)
t > 2π
(a)
Find the Laplace transform F(s) =
ℒ { f (t)} of f (t).
(b)
Find the Laplace transform Y(s) =
ℒ {y(t)} of the solution y(t)
of the above IVP.
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