Laplace transform for differential equation

  • #1

Homework Statement


use laplace transforms to solve the differential equation
y"+2y'+17y = 1

Homework Equations


Initial conditions are
y(0) = 0
y'(0) = 0

The Attempt at a Solution


so it converts to Y(s) (s^2+2s+17) = 1/s
which then ends up as;
Y(s) = 1/s*1/(s^2+2s+17)

i know i need to invert the equation from laplace again but not sure how.

Im not sure if im on the right track by re-arranging to 1/s((s+1)^2+4^2)?

Still getting used to spotting laplace eqns.
and help would be much appreciated.
 

Answers and Replies

  • #2
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Perhaps it would be easier to do two simple inverse Laplace transforms instead of one complicated one. Can you break that expression into a sum of two fractions using partial fraction decomposition?
 
  • #3
lurflurf
Homework Helper
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^complete the square
s^2+2s+17=(s+a)^2+b^2
what are a and b?
 
  • #4
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Im not sure if im on the right track by re-arranging to 1/s((s+1)^2+4^2)?
^complete the square
s^2+2s+17=(s+a)^2+b^2
what are a and b?
The OP already has that figured out - a = 1 and b = 4

Perhaps it would be easier to do two simple inverse Laplace transforms instead of one complicated one. Can you break that expression into a sum of two fractions using partial fraction decomposition?
@welcomereef, yes you are on the right track. axmls's advice is what I would follow if I were doing this problem.
 
  • #5
STEMucator
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The equation in latex:

$$Y(s) = \frac{1}{s} \frac{1}{s^2+2s+17}$$

You could always go the good old convolution theorem route:

$$f(t) \star g(t) = \Lagr^{-1} \{F(s)\} \star \Lagr^{-1} \{G(s)\} = \Lagr^{-1}\{F(s) G(s)\}$$

Then simply using the fact that:

$$f(t) \star g(t) = \int_0^t f(t-x) g(x) \space dx$$

You should get the same answer as any other method.
 
  • #6
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The equation in latex:

$$Y(s) = \frac{1}{s} \frac{1}{s^2+2s+17}$$

You could always go the good old convolution theorem route:

$$f(t) \star g(t) = \Lagr^{-1} \{F(s)\} \star \Lagr^{-1} \{G(s)\} = \Lagr^{-1}\{F(s) G(s)\}$$
Do you mean ##\mathcal{L}^{-1}## in your work above?

The OP might not be aware of convolutions at this point. In any case, I think that decomposing the fraction above is quite a bit simpler.
Zondrina said:
Then simply using the fact that:

$$f(t) \star g(t) = \int_0^t f(t-x) g(x) \space dx$$

You should get the same answer as any other method.
 
  • #7
STEMucator
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Do you mean ##\mathcal{L}^{-1}## in your work above?

The OP might not be aware of convolutions at this point. In any case, I think that decomposing the fraction above is quite a bit simpler.
Yes indeed I did. For some reason \Lagr was displaying properly yesterday, but not today by the looks of it. Guess I have to go with \mathcal{L}.

I was also demonstrating there's more than one way to skin this problem, albeit a bit more tedious. I just thought it to be interesting.
 
  • #8
Ray Vickson
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Homework Statement


use laplace transforms to solve the differential equation
y"+2y'+17y = 1

Homework Equations


Initial conditions are
y(0) = 0
y'(0) = 0

The Attempt at a Solution


so it converts to Y(s) (s^2+2s+17) = 1/s
which then ends up as;
Y(s) = 1/s*1/(s^2+2s+17)

i know i need to invert the equation from laplace again but not sure how.

Im not sure if im on the right track by re-arranging to 1/s((s+1)^2+4^2)?

Still getting used to spotting laplace eqns.
and help would be much appreciated.
Do a partial fraction expansion of Y(s), just the way you would do it if you were integrating instead of inverting.
Do you mean ##\mathcal{L}^{-1}## in your work above?

The OP might not be aware of convolutions at this point. In any case, I think that decomposing the fraction above is quite a bit simpler.
He can also use the standard result (found in many Laplace transform tables) that
[tex] f(t)\: \leftrightarrow\: F(s) \; \Longrightarrow \; \int_0^t f(\tau) \, d \tau \: \leftrightarrow \frac{F(s)}{s} [/tex]

Of course, this follows right away from the "convolution" result, but does not require it, and maybe is even known before one has ever heard of convolution. It also follows more-or-less directly from
[tex] g(t)\: \leftrightarrow \: G(s) \:\Longrightarrow \: g'(t) \: \leftrightarrow \: s G(s) - g(0+),[/tex]
just by setting ##g(t) = \int_0^t f(\tau) \, d \tau##.
 
Last edited:
  • #9
rude man
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Homework Statement


use laplace transforms to solve the differential equation
y"+2y'+17y = 1

Homework Equations


Initial conditions are
y(0) = 0
y'(0) = 0

The Attempt at a Solution


so it converts to Y(s) (s^2+2s+17) = 1/s
which then ends up as;
Y(s) = 1/s*1/(s^2+2s+17)

i know i need to invert the equation from laplace again but not sure how.

Im not sure if im on the right track by re-arranging to 1/s((s+1)^2+4^2)?
.
Yes you are.
Use two theorems:
1. If g(t) ↔ G(s), then G(s+a) ↔ e-atg(t).
That gives you a very simple G(s+a) = 1/((s+1)^2+4^2) to invert.
So now Y(s) = G(s+a)/s.
Then, to take care of the s, consider
If f(t) ↔ F(s), then
(1/s)F(s) ↔ ∫f(t')dt' with limits 0 and t.
 

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