# Use Laplace Transforms to solve (help)

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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

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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#2

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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

**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

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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(Original post by

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.

**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.

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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#4

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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

**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

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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(Original post by

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

**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

t = 0

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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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#8

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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

**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

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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(Original post by

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?

**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?

L^-1 = `(e^4t)t - e^2t + e^4t

Im doing it on my own the course

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#10

(Original post by

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

**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

There are quite a lot of examples of using laplace for solving linear odes.

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(Original post by

So that's your answer for x(t).

There are quite a lot of examples of using laplace for solving linear odes.

**mqb2766**)So that's your answer for x(t).

There are quite a lot of examples of using laplace for solving linear odes.

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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