Use Laplace Transforms to solve (help)

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mlm1234
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Use Laplace Transforms to solve the following equation;
𝒅x/dt - 4x = 2e^2t + e^4t.
𝒙 = 𝟎
𝒕 = 𝟎

1) L(dx/dt) - 4L(x) = 2L(e^2t) + L(e^4t)

2) (sX (s) - X(0)) - 4X = 2/(s - 2) + 1/(s - 4)

3) (s - 4)X (s) = (3s - 10)/(s - 2) + 1/(s - 4)

4)

https://tutorial.math.lamar.edu/clas...ace_Table.aspx



really dont know. dont have tutor to walk me through it
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mqb2766
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(Original post by mlm1234)
Use Laplace Transforms to solve the following equation;
𝒅x/dt - 4x = 2e^2t + e^4t.
𝒙 = 𝟎
𝒕 = 𝟎

1) L(dx/dt) - 4L(x) = 2L(e^2t) + L(e^4t)

2) (sX (s) - X(0)) - 4X = 2/(s - 2) + 1/(s - 4)

3) (s - 4)X (s) = (3s - 10)/(s - 2) + 1/(s - 4)

4)

https://tutorial.math.lamar.edu/clas...ace_Table.aspx



really dont know. dont have tutor to walk me through it
You've started correct, although i m not sure what the initial condition is or why
2/(s-2) becomes (3s-10)/(s-2)?

But divide through by s-4 and take inverse laplace to go from X(s) to x(t), once the previous points are sorted.
Last edited by mqb2766; 1 month ago
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mlm1234
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(Original post by mqb2766)
You've started correct, although i m not sure what the initial condition is or why
2/(s-2) becomes (3s-10)/(s-2)?

But divide through by s-4 and take inverse laplace to go from X(s) to x(t), once the previous points are sorted.
Hi i didnt know what to do when getting to the = 2/(s - 2) + 1/(s - 4)


and the whole question i got was
Use Laplace Transforms to solve the following equation;
𝒅x/dt - 4x = 2e^2t + e^4t.
𝒙 = 𝟎
𝒕 = 𝟎
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mqb2766
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(Original post by mlm1234)
Hi i didnt know what to do when getting to the = 2/(s - 2) + 1/(s - 4)


and the whole question i got was
Use Laplace Transforms to solve the following equation;
𝒅x/dt - 4x = 2e^2t + e^4t.
𝒙 = 𝟎
𝒕 = 𝟎
the last two lines are not showing on my browser. Is x(0) given?

You want to get to something like
X (s) = 2/(s - 2)(s-4) + 1/(s - 4)^2 + X(0)/(s-4)
Then invert
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mlm1234
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(Original post by mqb2766)
the last two lines are not showing on my browser. Is x(0) given?

You want to get to something like
x (s) = 2/(s - 2)(s-4) + 1/(s - 4)^2 + x(0)/(s-4)
then invert
x = 0
t = 0
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mqb2766
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(Original post by mlm1234)
x = 0
t = 0
In that case, zero initial condition, what do you get for
X(s) = ...
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mlm1234
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(Original post by mqb2766)
In that case, zero initial condition, what do you get for
X(s) = ...
sorry for being thick

1) L(dx/dt) - 4L(x) = 2L(e^2t) + L(e^4t)

2) (sX (s) - X(0)) - 4X = 2/(s - 2) + 1/(s - 4)


3) (s - 4)X (s) = 2/(s - 2) + 1/(s - 4)

4)
workings
(s-4) x 1/(s-4) = 1/(s-4)^2
(s-4) x 2/(s-2) = 2/(s-2)(s-4)

X = 1/(s-4)^2 + 2/(s-2)(s-4)

How did you get X(0)/(s-4) for the RHS

Do i just plug in 0 into the S to find the value
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mqb2766
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(Original post by mlm1234)
sorry for being thick

1) L(dx/dt) - 4L(x) = 2L(e^2t) + L(e^4t)

2) (sX (s) - X(0)) - 4X = 2/(s - 2) + 1/(s - 4)


3) (s - 4)X (s) = 2/(s - 2) + 1/(s - 4)

4)
workings
(s-4) x 1/(s-4) = 1/(s-4)^2
(s-4) x 2/(s-2) = 2/(s-2)(s-4)

X = 1/(s-4)^2 + 2/(s-2)(s-4)

How did you get X(0)/(s-4) for the RHS

Do i just plug in 0 into the S to find the value
There is no initial condition (it's zero).. So
X(s) = 1/(s-4)^2 + 2/(s-2)(s-4)
So use the tables to invert both terms. You'll need to use partial fractions on the second term to split it up first.

You must have some examp!es in your notes or just Google it?
Last edited by mqb2766; 1 month ago
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mlm1234
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(Original post by mqb2766)
There is no initial condition (it's zero).. So
X(s) = 1/(s-4)^2 + 2/(s-2)(s-4)
So use the tables to invert both terms. You'll need to use partial fractions on the second term to split it up first.

You must have some examp!es in your notes or just Google it?
X(s) = 1/(s-4)^2 + 2/(s-2)(s-4)
L^-1 = `(e^4t)t - e^2t + e^4t

Im doing it on my own the course
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mqb2766
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(Original post by mlm1234)
X(s) = 1/(s-4)^2 + 2/(s-2)(s-4)
L^-1 = `(e^4t)t - e^2t + e^4t

Im doing it on my own the course
So that's your answer for x(t).
There are quite a lot of examples of using laplace for solving linear odes.
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mlm1234
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(Original post by mqb2766)
So that's your answer for x(t).
There are quite a lot of examples of using laplace for solving linear odes.
Thank you so much for for the help and sorry for any bother. life saver
Know done and found (X(t) = (e^4t)t - e^2t + e^4t) it i can use that to under stand others.


just need walking through it to understand.
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