# Laplace transform for differential equation

• welcomereef
In summary, using the Laplace transforms, solving for Y(s) from y(0), y'(0), and the initial condition y(0), results in the following: Y(s) = 1/s*1/(s^2+2s+17).

## 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 I am 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.

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?

^complete the square
s^2+2s+17=(s+a)^2+b^2
what are a and b?

welcomereef said:
Im not sure if I am on the right track by re-arranging to 1/s((s+1)^2+4^2)?

lurflurf said:
^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

axmls said:
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.

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.

Zondrina said:
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.

Mark44 said:
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.

welcomereef said:

## 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 I am 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.
Mark44 said:
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
$$f(t)\: \leftrightarrow\: F(s) \; \Longrightarrow \; \int_0^t f(\tau) \, d \tau \: \leftrightarrow \frac{F(s)}{s}$$

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
$$g(t)\: \leftrightarrow \: G(s) \:\Longrightarrow \: g'(t) \: \leftrightarrow \: s G(s) - g(0+),$$
just by setting ##g(t) = \int_0^t f(\tau) \, d \tau##.

Last edited:
welcomereef said:

## 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 I am 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.