# Laplace Transforms and differential equations

• Aug 15th 2010, 05:07 AM
harrisonjane70
Laplace Transforms and differential equations
I am new to Laplace transforms and have started from the beginning with learning it.
I am troubled by the process of solving differential equations.

I have the problem:

dx/dt -4x = 8

I have rearranged to get sx - 2 - 4x = 8/s

now not really sure where to go from here. Can someone please show me?
• Aug 15th 2010, 05:18 AM
Prove It
No need for Laplace Transforms, this is first order linear, so you can use the integrating factor method...

$\displaystyle \frac{dx}{dt} - 4x = 8$

The integrating factor is $\displaystyle e^{\int{-4\,dt}} = e^{-4t}$, so multiplying through by the integrating factor gives

$\displaystyle e^{-4t}\frac{dx}{dt} - 4x\,e^{-4t} = 8e^{-4t}$

$\displaystyle \frac{d}{dt}(x\,e^{-4t}) = 8e^{-4t}$

$\displaystyle x\,e^{-4t} = \int{8e^{-4t}\,dt}$

$\displaystyle x\,e^{-4t} = -2e^{-4t} + C$

$\displaystyle x = -2 + Ce^{4t}$.
• Aug 15th 2010, 05:19 AM
CaptainBlack
Quote:

Originally Posted by harrisonjane70
I am new to Laplace transforms and have started from the beginning with learning it.
I am troubled by the process of solving differential equations.

I have the problem:

dx/dt -4x = 8

I have rearranged to get sx - 2 - 4x = 8/s

now not really sure where to go from here. Can someone please show me?

Use an Upper case $\displaystyle X(s)$ to denote the LT of $\displaystyle x(t)$.

Then you have:

$\displaystyle X(s) (s-4)=\dfrac{8}{s}+2$

so:

$\displaystyle X(s)=\dfrac{8+2s}{s(s-4)}$

Which you find the ILT of by the usual methods (partial fractions and table look up)

CB
• Aug 15th 2010, 05:35 AM
harrisonjane70
how about dx/dt + 2x = 10e^3t

rearranged to X(s)(s + 2) - 6 = 10 / s - 3
• Aug 15th 2010, 05:39 AM
Prove It
Again, first order linear. No need for Laplace Transforms...

The integrating factor is $\displaystyle e^{\int{2\,dt}} = e^{2t}$.

So multiplying through by the integrating factor gives

$\displaystyle e^{2t}\,\frac{dx}{dt} + 2x\,e^{2t} = 10e^{5t}$

$\displaystyle \frac{d}{dt}(x\,e^{2t}) = 10e^{5t}$

$\displaystyle x\,e^{2t} = \int{10e^{5t}\,dt}$

$\displaystyle x\,e^{2t} = 2e^{5t} + C$

$\displaystyle x = 2e^{3t} + Ce^{-2t}$.
• Aug 15th 2010, 05:43 AM
harrisonjane70
Is it possible to see it with Laplace Transforms? I'm looking at examples in a text book.
• Aug 15th 2010, 06:49 AM
harrisonjane70
transposing to obtain X somehow gets (6s - 8) / (s + 2)(s - 3)???

Can someone explain why the 10 has become -8?
• Aug 15th 2010, 08:26 AM
Krizalid
Prove It, he's meant to do it by applying LT according his(er) professor.

i've found many threads here when people say "integrate by using trig. substitution," then i appear and say "no need for trig. sub." but then one realises that they are learning how to use a new technique on solving a problem.
• Aug 15th 2010, 08:41 AM
harrisonjane70
yeah I'm going into year 2 Electrical engineering in october and I will be studying this in one of my modules along with fourier series and matrix algebra. At the moment I'm just looking at the material trying to see what Im up against as Im not the most mathematical person.
• Aug 15th 2010, 10:10 AM
CaptainBlack
Quote:

Originally Posted by harrisonjane70
transposing to obtain X somehow gets (6s - 8) / (s + 2)(s - 3)???

Can someone explain why the 10 has become -8?

It is called algebra:

$\displaystyle X(s)(s+2) - 6=\dfrac{10}{s-3}$

becomes:

$\displaystyle X(s)(s+2)=\dfrac{10}{s-3}+6=\dfrac{10+6(s-3)}{s-3}$

etc

CB
• Aug 15th 2010, 12:33 PM
mr fantastic
This question was asked by the same person using a different screen name here: http://www.mathhelpforum.com/math-he...ms-153657.html

Interesting that the same responses have been given ....