# Math Help - Laplace Transforms to solve Initial-Value problem

1. ## Laplace Transforms to solve Initial-Value problem

Use Laplace transforms to solve the initial-value problem,

$y^{''} + 4y^{'} + 4y = e^{-t}(sint + cost), y(0) = 0, y'(0) = 0.$

See figure(s) attached for my work.

I think I started this problem off correctly, however when it comes time to take the inverse laplace transform to obtain y(t) I think I may have taken the wrong route.

Should I be using partial fractions for this or is there another way? (We are a given a table of Laplace transforms that we can readily use without proof)

Let me know what you think.

Thanks again!

3. Originally Posted by dwsmith
I haven't found one. I stopped after I realized how ugly the partial fraction equations were. (We usually don't get things that ugly)

I've got a feeling I've made a mistake along the way of obtaining my equations for partial fractions but I'm not entirely sure.

4. By using partial fractions, you will obtain equations that are easier to do the inverse Laplace Transform

5. Originally Posted by jegues
Use Laplace transforms to solve the initial-value problem,

$y^{''} + 4y^{'} = e^{-t}(sint + cost), y(0) = 0, y'(0) = 0.$

See figure(s) attached for my work.

I think I started this problem off correctly, however when it comes time to take the inverse laplace transform to obtain y(t) I think I may have taken the wrong route.

Should I be using partial fractions for this or is there another way? (We are a given a table of Laplace transforms that we can readily use without proof)

Let me know what you think.

Thanks again!
Take the Laplace transform gives

$\displaystyle s^2Y-4sY=\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}$

This gives

$\displaystyle Y= \frac{1}{s(s-4)[(s+1)^2+1]} + \frac{s+1}{s(s-4)[(s+1)^2+1]}$

Now comes the hard part, inverting the transform.

First we will use the convolution theorem twice to invert each Fraction. Lets focus on

$\displaystyle \frac{1}{(s-4)}\cdot\frac{1}{[(s+1)^2+1]}$

The inverse tranform of this is

$\displaystyle \int_{0}^{t}e^{4(t-\tau)}e^{-\tau}\sin(\tau)d\tau = e^{4t}\int_{0}^{t}e^{-5\tau}\sin(\tau)d\tau=e^{4t}\left( \frac{-e^{-5\tau}}{26}\left[ 5\sin(\tau)+\cos(\tau)\right]\right)\bigg|_{0}^{t}=$
$\displaystyle \frac{1}{26}\left( e^{4t}-5e^{-t}\sin(t)-e^{-t}\cos(t)\right)$

Now we are half way there. Using this we can use the convolution theorem again to invert

$\frac{1}{s}\cdot \left( \frac{1}{(s-4)}\cdot\frac{1}{[(s+1)^2+1]}\right)$

This gives

$\displaystyle \frac{1}{26}\int_{0}^{t}\left( e^{4\tau}-5e^{-\tau}\sin(\tau)-e^{-\tau}\cos(\tau)\right)d\tau=-\frac{1}{8}+\frac{1}{104}e^{4t}+\frac{3}{26}e^{-t}\cos(t)+\frac{1}{14}e^{-t}\sin(t)$

Here is half now you can try the other one.

6. EmptySet, the problem posted is different from the image. It is suppose to be $\displaystyle y''+4y'+4y=...$

7. Originally Posted by dwsmith
EmptySet, the problem posted is different from the image. It is suppose to be $\displaystyle y''+4y'+4y=...$
Whoops! Sorry about that EmptySet, I edited the OP now to reflect the correct problem statement.

Can you give it another crack?

8. Originally Posted by jegues
Whoops! Sorry about that EmptySet, I edited the OP now to reflect the correct problem statement.

Can you give it another crack?
This principle is the same If you have two Laplace transforms

$\displaystyle \mathcal{L}^{-1}(F(s)G(s))=\int_{0}^{t}f(t-\tau)g(\tau)d\tau$

$\displaystyle \frac{1}{(s+2)^2}\left(\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}\right)$

So we need to use the convolution theorem twice again starting with

$\displaystyle \frac{1}{(s+2)}\left(\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}\right)$

This gives

$\int_{0}^{t}e^{-2(t-\tau)}\left( e^{-\tau}(\sin(\tau)+\cos(\tau)\right)d\tau=e^{-t}\sin(t)$

Now use the convolution theorem again on

$\displaystyle \frac{1}{(s+2)}\left[\frac{1}{(s+2)}\left(\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}\right)\right]$

This gives

$\displaystyle \int_{0}^{t}e^{-2(t-\tau)}e^{-\tau}\sin(\tau)d\tau=\frac{e^{-2t}}{2}+\frac{e^{-t}\sin(t)}{2}-\frac{e^{-t}\cos(t)}{2}$

9. Originally Posted by TheEmptySet
This principle is the same If you have two Laplace transforms

$\displaystyle \mathcal{L}^{-1}(F(s)G(s))=\int_{0}^{t}f(t-\tau)g(\tau)d\tau$

$\displaystyle \frac{1}{(s+2)^2}\left(\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}\right)$

So we need to use the convolution theorem twice again starting with

$\displaystyle \frac{1}{(s+2)}\left(\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}\right)$

This gives

$\int_{0}^{t}e^{-2(t-\tau)}\left( e^{-\tau}(\sin(\tau)+\cos(\tau)\right)d\tau=e^{-t}\sin(t)$

Now use the convolution theorem again on

$\displaystyle \frac{1}{(s+2)}\left[\frac{1}{(s+2)}\left(\frac{1}{(s+1)^2+1} + \frac{s+1}{(s+1)^2+1}\right)\right]$

This gives

$\displaystyle \int_{0}^{t}e^{-2(t-\tau)}e^{-\tau}\sin(\tau)d\tau=\frac{e^{-2t}}{2}+\frac{e^{-t}\sin(t)}{2}-\frac{e^{-t}\cos(t)}{2}$
Ah, we didn't cover convolution theorem in my course, so maybe this question is out of our reach.

Is there another suitable way of getting this inverse?

10. Maybe I'm missing something here, but don't you get

$Y(s+2)^{2}=\dfrac{s+2}{(s+1)^{2}+1},$ which implies

$Y=\dfrac{1}{(s+2)((s+1)^{2}+1)}?$ Then postulate the partial fraction expansion

$Y=\dfrac{A}{s+2}+\dfrac{Bs+C}{(s+1)^{2}+1},$

add the fractions, set the numerators equal, solve the system of three equations in three unknowns, etc. It's not all that bad, I don't think.

11. I agree with Ackbeet. I get the same thing he does.

12. Originally Posted by jegues
Use Laplace transforms to solve the initial-value problem,

$y^{''} + 4y^{'} + 4y = e^{-t}(sint + cost), y(0) = 0, y'(0) = 0.$

See figure(s) attached for my work.

I think I started this problem off correctly, however when it comes time to take the inverse laplace transform to obtain y(t) I think I may have taken the wrong route.

Should I be using partial fractions for this or is there another way? (We are a given a table of Laplace transforms that we can readily use without proof)

Let me know what you think.

Thanks again!
You are on the right track but try this. It will make it easier to find your constants

using what ackbeet said just modify a little

$Y=\dfrac{A}{s+2}+\dfrac{B(s+1)+C}{(s+1)^{2}+1}$

so

$s+2 = A[(s+1)^2+1] + (s+2)[B(s+1) + C]$

now instead of expanding you can force a solution.

lets try s = -2

$0 = A[(-2+1)^2 + 1] +(-2 +2)[B(-2+1) + C]$

$0 = 2A$

so

A = 0

Now to find the other constants you can plug in 2 different values for s and make a system of linear equations and solve.

13. Originally Posted by 11rdc11
You are on the right track but try this. It will make it easier to find your constants

using what ackbeet said just modify a little

$Y=\dfrac{A}{s+2}+\dfrac{B(s+1)+C}{(s+1)^{2}+1}$

so

$s+2 = A[(s+1)^2+1] + (s+2)[B(s+1) + C]$

now instead of expanding you can force a solution.

lets try s = -2

$0 = A[(-2+1)^2 + 1] +(-2 +2)[B(-2+1) + C]$

$0 = 2A$

so

A = 0

Now to find the other constants you can plug in 2 different values for s and make a system of linear equations and solve.
I think you need to have

$\displaystyle \frac{A}{(s+2)}+\frac{B}{(s+2)^2}$

14. Originally Posted by TheEmptySet
I think you need to have

$\displaystyle \frac{A}{(s+2)}+\frac{B}{(s+2)^2}$

Yes I think I made a mistake and you are correct. My solution of $y = e^{-t}\sin{t}$ does not work

15. Are Laplace Transforms really necessary here? This is second-order linear constant coefficient non-homogeneous differential equation.

Homogeneous solution:

The characteristic equation is $\displaystyle m^2 + 4m + 4 = 0$

$\displaystyle (m + 2)^2 = 0$

$\displaystyle m +2 = 0$

$\displaystyle m = -2$ with multiplicity $\displaystyle 2$.

Since the root is repeated, the homogeneous solution is $\displaystyle y = C_1 e^{-2t} + C_2t\,e^{-2t}$.

Non-homogeneous solution: Assume a solution of the form $\displaystyle y = Ae^{-t}\sin{t} + Be^{-t}\cos{t}$.

Then $\displaystyle \frac{dy}{dx} = Ae^{-t}\cos{t} - Ae^{-t}\sin{t} - Be^{-t}\sin{t} - Be^{-t}\cos{t} = -(A+B)e^{-t}\sin{t} + (A-B)e^{-t}\cos{t}$

and $\displaystyle \frac{d^2y}{dx^2} = (A+B)e^{-t}\sin{t} - (A+B)e^{-t}\cos{t} + (B-A)e^{-t}\cos{t} + (B-A)e^{-t}\sin{t} = 2Be^{-t}\sin{t} -2Ae^{-t}\cos{t}$.

Substituting into your original DE gives

$\displaystyle 2Be^{-t}\sin{t} - 2Ae^{-t}\cos{t} - 4(A+B)e^{-t}\sin{t} + 4(A-B)e^{-t}\cos{t} + 4Ae^{-t}\sin{t} + 4Be^{-t}\cos{t} = e^{-t}\sin{t} + e^{-t}\cos{t}$

$\displaystyle [2B-4(A+B)+4A]e^{-t}\sin{t} + [-2A + 4(A-B) + 4B]e^{-t}\cos{t} = e^{-t}\sin{t} + e^{-t}\cos{t}$

$\displaystyle -2Be^{-t}\sin{t} + 2Ae^{-t}\cos{t} = e^{-t}\sin{t} + e^{-t}\cos{t}$.

Equating like coefficients gives $\displaystyle -2B = 1$ and $\displaystyle 2A = 1$.

So $\displaystyle A = \frac{1}{2}$ and $\displaystyle B = -\frac{1}{2}$.

$\displaystyle y = \frac{1}{2}e^{-t}\sin{t} - \frac{1}{2}e^{-t}\cos{t}$.
$\displaystyle y = C_1e^{-2t} + C_2t\,e^{-2t} + \frac{1}{2}e^{-t}\sin{t} - \frac{1}{2}e^{-t}\cos{t}$.