Laplace transform for differential equation

[welcomereef]

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.

 

[axmls]

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?

 

[lurflurf]

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

 

[Mark44]

Im not sure if I am 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.

 

[STEMucator]

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.

 

[Mark44]

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.

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.

 

[STEMucator]

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.

 

[Ray Vickson]

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.

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 $g(t) = \int_0^t f(\tau) \, d \tau$.

 

[rude man]

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.