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

[aeroguy2008]

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.

 

[aeroguy2008]

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

 

[aeroguy2008]

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

 

[Defennder]

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

 

[aeroguy2008]

Defennder thanks for pointing that out. So now I have the following:

 

[aeroguy2008]

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?

 

[Defennder]

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.

 

[aeroguy2008]

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.

Thanks for letting me know...I hope this works:)
http://img57.imageshack.us/my.php?image=mathprobkv7.jpg"

 

[Defennder]

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.

 

[aeroguy2008]

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]

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.

 

[aeroguy2008]

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

 

[Defennder]

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

 

[aeroguy2008]

I have the following...

http://img71.imageshack.us/img71/4016/mathprob1na5.jpg"

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...

 

[aeroguy2008]

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]

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.

 

[aeroguy2008]

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?

 

[aeroguy2008]

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?

 

[aeroguy2008]

Can't anyone shed more light on this problem?

 

[Defennder]

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:

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

 

[aeroguy2008]

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 going to 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.

 

[Defennder]

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.

 

[aeroguy2008]

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]

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.

 

[aeroguy2008]

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 can't get an exact fraction I get this to 1.05...how do you get exact fractions?

 

[Defennder]

$$u(t-a) = \begin{array}{cc} 0 \ \mbox{for} \ t<a\\1 \ \mbox{for} \ t>a \end{array}$$

So we have t=14pi/9, so what is value of u(t-pi)? Get that value, put it into the y(t) and some term will drop out and the resulting expression can be evaluated using a trigo table.

 

[aeroguy2008]

$$u(t-a) = \begin{array}{cc} 0 \ \mbox{for} \ t<a\\1 \ \mbox{for} \ t>a \end{array}$$

So we have t=14pi/9, so what is value of u(t-pi)? Get that value, put it into the y(t) and some term will drop out and the resulting expression can be evaluated using a trigo table.

Yes now I understand what you mean. The answer to this problem is thus a/b=-1/2 :) This problem is now officially solved :D Defennder, thank you so much for your help. I learned alot!

 

[HallsofIvy]

This is a perfect example of why I think Laplace Transforms are over emphasized in elementary differential equations courses. Laplace Transforms can only be applied to problems involving non-homogeneous linear differential equations with constant coefficients for which there are other, simpler methods. I suppose engineers like to be able to think they can just look up solutions in tables of inverse transforms.

This problem is an ideal candidate for "variation of parameters". Two independent solutions to the homogeneous equation y"+ 9y= 0 are cos(3t) and sin(3t). Now look for a solution to the entire equation of the form y(t)= u(t)cos(3t)+ v(t)sin(3t) where u(t) and v(t) are differentiable functions to be determined. Of course, since there are an infinite number of solutions to that equation, there are an infinite number of such functions u and v.

y'= u'cos(3t)- 3usin(3t)+ v'sin(3t)+ 3vcos(3t). Since there are an infinite number of workable solutions, we narrow the search (and simplify the calculations) by requiring that u'cos(3t)+ v' sin(3t)= 0. With that requirement, y'= -3u sin(3t)+ 3v cos(3t). Differentiating again, y"= -3u' sin(3t)- 9u cos(3t)+ 3v' cos(3t)- 9v sin(3t). Then y"+ 9y= -3u' sin(3t)- 9u cos(3t)+ 3v' cos(3t)- 9v sin(3t)+ 9u cos(3t)+ 9v sin(3t)= -3u' sin(3t)+ 3v' cos(3t).

We must have $-3u' sin(3t)+ 3v' cos(3t)= \delta(t- \pi)$ and u' cos(3t)+ v' sin(3t)= 0.

We can solve those two equations algebraically for u' and v':
$$u'= -\frac{1}{3}sin(3t)\delta(t- \pi)$$
and
$$v'= \frac{1}{3}cos(3t)\delta(t- \pi)$$

u and v are, of course, the integrals of those. Since the integral of $f(x)\delta(t- \pi)$ is 0 if the interval of integration does NOT contain $\pi$ and $f(\pi)$ if it does, the integral, from $-\infty$ to t of $(-1/3)sin(3t)\delta(t-\pi)$ is 0 if $t< \pi$ and $(-1/3)sin(3\pi)= 0$ if $t\ge 3\pi$. In other words, u(t) is identically 0.

v(t) is 0 if $t< \pi$ and $(1/3)cos(3\pi)= -1/3$ if $t\ge \pi$. That is, $v(t)= (-1/3)H(t- \pi)$ where $H(t- \pi)$ is the Heaviside unit step function.

The general solution to that differential equation is
$$y(t)= C cos(3t)+ D sin(3t)- (1/3)sin(3t)H(t-\pi)$$