# Math Help - Differential Equations (Laplace Transforms)

1. ## 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.

2. 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

$\int_0^{t}e^{-2\tau}\sin(\tau) d\tau$

3. 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

$x(t)=\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-2\tau}\sin(3\tau)d\tau$

4. 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.

5. 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

$\frac{1}{2}\sin(2t)$

6. 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

7. 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

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

I'm really rusty at this stuff.

9. ## number 2

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

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

10. ## #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

$x(t)=\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-\tau}d\tau-\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-4\tau}d\tau$

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

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?

12. 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.

13. 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.

14. 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.

15. 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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