# Solve y''+9y=delta(t-pi)

Solve the differential equation

y''+9y=delta(t-pi)

that fulfills the initial condition y(0)=y(0)=1. Answer by giving the value for y((14*pi)/9). The answer can be given by a fraction a/b.

## The Attempt at a Solution

I will submit my attempt for a solution as an attachment shortly.

I am now at the stage where I need to get the inverse transforms but I don't remember how...

#### Attachments

• mathproblemxx.jpg
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I can't figure out how to get the inverse transforms...any ideas?

Defennder
Homework Helper
There is a mistake on the 3rd line. $$sY(0)$$ is not 1. It's just s. Also $$s^2 + 9$$ is not $$(s+3)^2$$.

I just don't understand how these three fractions can give me y at the end. Is the approach of inverse transform incorrect? Cause, just by looking at s/(s^2+9) I can transform that to cos(3t) but what do I do with the rest. How can these three "add up" to give me the fraction a/b?

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Defennder
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Your attachment hasn't been approved yet. Try hosting it on imageshack.us and linking the picture. That way you won't have to wait for approval from the mentors.

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Defennder
Homework Helper
So far so good. I got the same answer. To inverse-Laplace transform the RHS into a function of t again, you'll need the laplace transform of:

$$L[\cos (\omega t)]$$
$$L[\sin (\omega t)]$$
$$L[f(t-a)u(t-a)]$$ where u(t-a) is the unit step function or Heaviside function. Look this up and inverse-transform them individually.

The Laplace Transform of cos(at) is s/(s^2+a^2)
The Laplace Transform of sin(at) is a/(s^2+a^2)
and Laplace Transform of f(t-a)u(t-a) is exp(-Ts)F(s)

Now I will try to inverse-transform them individually.

Defennder
Homework Helper
The Laplace Transform of cos(at) is s/(s^2+a^2)
The Laplace Transform of sin(at) is a/(s^2+a^2)
and Laplace Transform of f(t-a)u(t-a) is exp(-Ts)F(s)

Now I will try to inverse-transform them individually.
All ok except this one. Probably a typo.

Oh yeah and Laplace Transform of f(t-a)u(t-a) is exp(-a*s)F(s)
is that correct now?

Defennder
Homework Helper
Yeah, it is. You should be able to solve for y now.

I have the following...

http://img71.imageshack.us/img71/4016/mathprob1na5.jpg" [Broken]

But I'm not sure about that Heaviside function term. I really don't know what to do with (s^2+9) in denominator? Isn't there a jump (discontinuity in Heaviside functions)? I have not y yet. If I add these three is that how I get y? What do I have to do about that F(s) that popped up? And do I just plug in t=((14*pi)/9) at the end when I get y? I am thankful for any suggestions...

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After some thinking about the Heaviside term I came to the conclusion that the last term in the attachment above should have an inverse Laplace transform of: u(t)f(t-pi)*(1/3)*sin(3t)

If this is correct (please object if it is not) as I add all three terms I should get:

y(t)=u(t)f(t-pi)*(1/3)*sin(3t)+(1/3)*sin(3t)+cos(3t)

According to my understanding I should now plug in t=((14*pi)/9) into y(t) as shown above:

y(t=((14*pi)/9))=u((14*pi)/9))f((14*pi)/9)-pi)*(1/3)*sin(3*(14*pi)/9))+(1/3)*sin(3*(14*pi)/9))+cos(3*(14*pi)/9))

Defennder
Homework Helper
After some thinking about the Heaviside term I came to the conclusion that the last term in the attachment above should have an inverse Laplace transform of: u(t)f(t-pi)*(1/3)*sin(3t)
This isn't right. Remember that both the Heaviside function and the function f(t) is shifted by -a. So u(t) is not it. f(t) here, which I take to be 1/3sin(3t) is also not shifted correctly.

This isn't right. Remember that both the Heaviside function and the function f(t) is shifted by -a. So u(t) is not it. f(t) here, which I take to be 1/3sin(3t) is also not shifted correctly.

I see...is this what you mean?

u(t-a)f(t-a)*(1/3)*sin(3t-a) Is the a-value for the shift pi?

If this is true, will this be the expression for y:

y=u(t-pi)f(t-pi)1/3sin(3t-pi)+1/3sin(3t)+cos(3t)

How will I evaluate the first term in the expression above numerically?

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I plugged in t=((14*pi)/9) into cos(3t) & 1/3sin(3t)...

cos(3t)=-0.5
1/3sin(3t)=-0.32827

I still don't know what to to about the heaviside-term. How can I evaluate that numerically? And how do I get a fraction a/b from all this? Any ideas or suggestions?

Can't anyone shed more light on this problem?

Defennder
Homework Helper
aeroguy2008 said:
u(t-a)f(t-a)*(1/3)*sin(3t-a)
That's not the inverse transform of $\frac{e^{-\pi s}}{s^2+9}$. Remember that f(t-a) is not some unknown function; it's just the inverse Laplace transform of $$\frac{1}{s^2+9}$$, which is shifted by -a. What this means is that in place of t in the original inverse transform of $$\frac{1}{s^2+9}$$ you now must write $t-a$. And after that is done, write u(t-a) next as a term multiplied to it.

As for the meaning of the Heaviside function u(t-a), note that all it does is ensure that the graph of f(t)u(t-a) would be 0 for t<a. For t>a, the graph would simply be f(t-a). You can treat u(t-a)f(t) as a shorthand for writing:

$$f(t)u(t-a) = \begin{array}{c}0 \ \mbox{for} \ t<a\\f(t) \ \mbox{for} \ t>a \end{array}$$

That's not the inverse transform of $\frac{e^{-\pi s}}{s^2+9}$. Remember that f(t-a) is not some unknown function; it's just the inverse Laplace transform of $$\frac{1}{s^2+9}$$, which is shifted by -a. What this means is that in place of t in the original inverse transform of $$\frac{1}{s^2+9}$$ you now must write $t-a$. And after that is done, write u(t-a) next as a term multiplied to it.

As for the meaning of the Heaviside function u(t-a), note that all it does is ensure that the graph of f(t)u(t-a) would be 0 for t<a. For t>a, the graph would simply be f(t-a). You can treat u(t-a)f(t) as a shorthand for writing:

$$f(t)u(t-a) = \begin{array}{c}0 \ \mbox{for} \ t<a\\f(t) \ \mbox{for} \ t>a \end{array}$$

Ok I have read your message 20 times now...and I'm still confused. I follow until the fact that f(t-a) is the inverse Laplace transform of 1/(s^2+9) shifted by -a. So then f(t-a)=1/3sin(3(t-a))u(t-a)? Is a=pi? How do I get the fraction a/b out of all this? I'm gonna plug in some numerics and see what I end up in my next post. Sorry you must think this is really easy but I am a little lost.

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Defennder
Homework Helper
So then f(t-a)=1/3sin(3(t-a))u(t-a)? Is a=pi? How do I get the fraction a/b out of all this?
Almost correct. f(t-a)u(t-a) is 1/3sin(3(t-a))u(t-a) which is in turn the inverse Laplace transform of $$\frac{e^{-\pi s}}{s^2+9}$$. Now combine all the three inverses into y(t) and find y(14pi/9). Remember how u(t-a)f(t) is supposed to be interpreted. Your final answer should be a fraction of integers.

So then my u(t-a)f(t) is equal to f(t) as Pi is smaller than (14*Pi)/9. Now I will only have:

t=((14*pi)/9)=4.886921906 into cos(3t) & 1/3sin(3t)...
cos(3t)=-0.5
1/3sin(3t)=-0.32827
f(t-a)=1.745329

y=1.745329-0.5-0.32827=0.917059 Is this correct?

Defennder
Homework Helper
For one thing you shouldn't use a decimal approximation to 14pi/9, because that would not allow you to express your answer in terms of a fraction. You have
$$y(t) = \frac{1}{3}\sin (3t) + \cos (3t) + \frac{1}{3}u(t-\pi)\sin(3t-3\pi)$$
which can be simplified by simplifying sin(3t-pi):
$$y(t) = \frac{1}{3}\sin (3t)(1-u(t-\pi)) + \cos (3t)$$

Now you're told to substitute t=14pi/9. So what happens to u(t-pi) and the corresponding expression? You don't need a calculator for this one. Just a trigo table will do.

For one thing you shouldn't use a decimal approximation to 14pi/9, because that would not allow you to express your answer in terms of a fraction. You have
$$y(t) = \frac{1}{3}\sin (3t) + \cos (3t) + \frac{1}{3}u(t-\pi)\sin(3t-3\pi)$$
which can be simplified by simplifying sin(3t-pi):
$$y(t) = \frac{1}{3}\sin (3t)(1-u(t-\pi)) + \cos (3t)$$

Now you're told to substitute t=14pi/9. So what happens to u(t-pi) and the corresponding expression? You don't need a calculator for this one. Just a trigo table will do.

Thanks for all the help. I would say u(t-pi) goes to zero? So I will be left with
y=1/3sin(t)+cos(3t) ... I cant get an exact fraction I get this to 1.05...how do you get exact fractions?