# Differential Equations (Laplace Transforms)

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• Apr 15th 2008, 05:46 PM
thedoge
Differential Equations (Laplace Transforms)
Use Laplace Transforms to solve these IVP (partial fractions):

1. x''+4'+8x=e^-t x(0)=x'(0)=0
I don't know how to work with the imaginary roots.

2. x''''-x=0 x(0)=1 x'(0)=x''(0)=x'''(0)=0

3. x''''+13x''+36x=0 x(0)=x''(0)=0 x'(0)=2 x'''(0)=-13

Apply the convolution theorem to find the inverse Laplace transforms for:
4. F(s) = 1/(s(s^2+4))

5. F(s) = 1/(s(s^2+4s+5))

Apply the convolution theorem to derive the indicated solution x(t) of the given DE w/ initial conditions x(0)=x'(0) = 0

6.x''+4x'+13x=f(t);
x(t)=1/3 int[0,t] f(t-tau)e^-2(tau)*sin(3*tau) dtau

Solve the IVP

7. mx''+cx'+kx=f(t);
x(0) = x'(0)=0

m=1
k=4
c=5
f(t)=1 if 0=<t=<2
f(t)=0 if t>=2

Help on any is useful.
• Apr 15th 2008, 06:04 PM
TheEmptySet
Quote:

Originally Posted by thedoge
Use Laplace Transforms to solve these IVP (partial fractions):

apply the convolution theorem to find the inverse Laplace transforms for:
F(s) = 1/(s(s^2+4))

F(s) = 1/(s(s^2+4s+5))

any help on any is useful.

for the first one

$\frac{1}{s(s^2+4)}=\underbrace{\frac{1}{s}}_{f(t)= 1} \cdot \underbrace{\frac{1}{s^2+4}}_{g(t)=\frac{1}{2}\sin (2t)}$

so the inverse transform is

$\int_{0}^{t}f(t-\tau)g(\tau)d \tau=\int_{0}^{t}\frac{1}{2}\sin(2\tau)d\tau=$
$-\frac{1}{4}\cos(2\tau)|_{0}^{t}=-\frac{1}{4}\cos(2t)+\frac{1}{4}$

for the next one

$\frac{1}{s(s^2+4s+5)}=\frac{1}{s[(s+2)^2+1}$

$\underbrace{\frac{1}{s}}_{1} \cdot \underbrace{\frac{f(t)=1}{(s+2)^2+1}}_{g(t)=e^{-2t}\sin(t)}$

so evaluate this integral

$\int_0^{t}e^{-2\tau}\sin(\tau) d\tau$
• Apr 15th 2008, 06:14 PM
TheEmptySet
Apply the convolution theorem to derive the indicated solution x(t) of the given DE w/ initial conditions x(0)=x'(0) = 0

x''+4x'+13x=f(t);
x(t)=1/3 int[0,t] f(t-tau)e^-2(tau)*sin(3*tau) dtau

taking the Laplace transform we get

$s^2X+4sX+13X=F(s) \iff (s^2+4s+13)X=F(s)$

$X=\frac{F(s)}{s^2+4s+13}=\underbrace{F(s)}_{f(t)} \cdot \underbrace{\frac{1}{(s+2)^2+9}}_{g(t)=\frac{1}{3} e^{-2t}\sin(3t)}$

by the convolution theorem we get

$x(t)=\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-2\tau}\sin(3\tau)d\tau$
• Apr 15th 2008, 06:19 PM
thedoge
Can you clarify how you got g(t)=1/2sin(2t)?

It's probably something simple, but I'm missing it.

I think it has to do with the imaginary roots. As I said, I'm not sure how to work with those.
• Apr 15th 2008, 06:31 PM
TheEmptySet
Quote:

Originally Posted by thedoge
Can you clarify how you got g(t)=1/2sin(2t)?

It's probably something simple, but I'm missing it.

There are few ways the most common is to memorize some common formulas and use them back wards. If you have taken a course in complex variables I will show you how to evaluate the contour integral :)

$\mathcal{L}(sin(at))=\frac{a}{s^2+a^2}$

so we want to use this in reverse so

$\frac{1}{s^2+a^2}=\frac{1}{2} \cdot \frac{2}{s^2+a^2}$

by using the above formula backwards we get

$\frac{1}{2}\sin(2t)$
• Apr 15th 2008, 06:37 PM
thedoge
Thanks again, that makes perfect sense.

I'm trying to decide between PDEs and Complex Variables. I think PDEs would be more useful in my field (atmospheric science), but I don't really know what the Complex Variables course is about.

So, for the first problem, I have it here:

$
X \left( s \right) = \left( s+1 \left( {s}^{2}+4\,s+8 \right)
\right) ^{-1}
$

With Roots: 2+-2i, what happens now?
I can tell there will be an e^-2t and a cos(2t) sin(2t), but im hazy for the steps to get to the final answer
• Apr 15th 2008, 07:10 PM
TheEmptySet
Quote:

Originally Posted by thedoge
Thanks again, that makes perfect sense.

I'm trying to decide between PDEs and Complex Variables. I think PDEs would be more useful in my field (atmospheric science), but I don't really know what the Complex Variables course is about.

So, for the first problem, I have it here:

$
X \left( s \right) = \left( s+1 \left( {s}^{2}+4\,s+8 \right)
\right) ^{-1}
$

With Roots: 2+-2i, what happens now?
I can tell there will be an e^-2t and a cos(2t) sin(2t), but im hazy for the steps to get to the final answer

The transform would be

$
s^2X+4sX+8X=\frac{1}{s+1}
$

solving for X gives

$X=\frac{1}{(s+1)(s^2+4s+8)}$

using partial fractions we get

$X=\frac{1}{(s+1)(s^2+4s+8)}=\frac{1}{5}\frac{1}{(s-1)}-\frac{1}{5}\frac{(s+3)}{s^2+4s+8}$

Lets look at 2nd term complete the square of the denominator

$-\frac{1}{5}\frac{(s+3)}{s^2+4s+8}= -\frac{1}{5} \left( \frac{(s+2)+1}{(s+2)^2+4} \right) =-\frac{1}{5}\left( \frac{s+2}{(s+2)^2+4}+\frac{1}{(s+2)^2+4}\right)$

now using the s-axis theorem backwards we get

$\mathcal{L}^{-1}\left( -\frac{1}{5}\left( \frac{s+2}{(s+2)^2+4}+\frac{1}{(s+2)^2+4}\right) \right) =-\frac{1}{5} \left( e^{-2t}\cos(2t)+e^{-2t}\frac{1}{2}\sin(2t)\right)$

You should be able to finish from here
• Apr 15th 2008, 09:00 PM
thedoge
I'd still appreciate seeing the procedure for 2, 3 and 7 if someone feels like it.

I'm really rusty at this stuff.
• Apr 15th 2008, 09:19 PM
TheEmptySet
number 2
Quote:

Originally Posted by thedoge
I'd still appreciate seeing the procedure for 2, 3 and 7 if someone feels like it.

I'm really rusty at this stuff.

$s^4X-s^3-X=\frac{1}{s} \iff X= \frac{s^4+1}{s(s^4-1)}$

by partial fractions we get

$X=\frac{1}{s}+\frac{1}{2}\frac{1}{(s-1)}+\frac{1}{2}\frac{1}{(s+1)}+\frac{s}{s^2+1}$

taking the inverse Transform we get

$x(t)=1+\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}+\cos(t)$

note that the following can also be written as

$\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}=\cosh(t)$
• Apr 15th 2008, 09:29 PM
TheEmptySet
#7
starting with

$x''+5x'+4x=f(t)$

taking the LaPlace transform we get

$s^2X+5sX+4X=F(s) \iff (s^2+5s+4)X=F(s)$

$X=\frac{F(s)}{(x+4)(x+1)}$

by partial fractions we get

$X=F(s)\frac{1}{3}\frac{1}{(s+1)}-F(s)\frac{1}{3}\frac{1}{(s+4)}$

using the convolution theorem we get

$x(t)=\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-\tau}d\tau-\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-4\tau}d\tau$
• Apr 16th 2008, 06:23 AM
thedoge
$s^4X-s^3-X=\frac{1}{s} \iff X= \frac{s^4+1}{s(s^4-1)}$

I'm really confused how you have $\frac{1}{s}$ and the $s^3$.

It's $=0$ and the equation should be $s^4X-X$, right?
• Apr 16th 2008, 06:35 AM
TheEmptySet
Quote:

Originally Posted by thedoge
$s^4X-s^3-X=\frac{1}{s} \iff X= \frac{s^4+1}{s(s^4-1)}$

I'm really confused how you have $\frac{1}{s}$ and the $s^3$.

It's $=0$ and the equation should be $s^4X-X$, right?

$
x''''-x=0 ;x(0)=1 x'(0)=x''(0)=x'''(0)=0

$

You need to remember the intial conditions

$\mathcal{L} (x^{(4)})=s^4X-s^3x(0)-s^2x'(0)-sx''(0)-x'''(0)=s^4X-s^3(1)-s^2(0)-s(0)-0=$

$s^4X-s^3$

in general we gave

$\mathcal{L}(x^{(n)})=s^nx(0)-s^{n-1}x'-...-sx^{(n-1)}-x^{n}

$

where $x^{(n)}$ is the nth derivative :)

You are right the right hand side should be zero, I wrote down a 1 by mistake.
• Apr 16th 2008, 06:44 AM
TheEmptySet
I should have gotten

$s^4X-s^3-X=0 \iff X=\frac{s^3}{(s^4-1)}$

by partial fractions I get...

$X=\frac{1}{4} \left( \frac{1}{s-1}+\frac{1}{s+1}\right)+\frac{1}{2}\frac{s}{s^2+1}$

Taking the inverse Laplace transform I get

$x(t)=\frac{1}{4}e^{t}+\frac{1}{4}e^{-t}+\frac{1}{2}\cos(t)$

or as before

$x(t)=\frac{1}{2}\cosh(t)+\frac{1}{2}\cos(t)$

sorry about solving the wrong problem.
• Apr 16th 2008, 07:03 AM
thedoge
Thank you very much! I think I finally understand these now:) You were able to clarify what I was missing. I did plenty on my own, but I wasn't sure how to deal with these few I posted.
• Apr 16th 2008, 09:07 AM
thedoge
on 3...

I have it to
$
X(s) = \frac{-13s^3+4s-36}{s^2(s^2+13)}
$

Now do I do partial fractions like this?
$
X(s) = \frac{Ax+B}{s^2} + \frac{Cx+D}{s^2+13}
$
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