# 2nd order of differential equations

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• Mar 18th 2006, 10:07 PM
jenjen
2nd order of differential equations
Hi, I am studying for my final on Monday, 20th and I came across 4 questions on the 2nd order of differential equations that I couldn’t do. :( Please help me.

Problem #1: For what values of c do solutions $\displaystyle y"+cy'+y = 0$ have infinitely many zeros and tend to zero as t goes to infinity (here we are considering y as a function of t).

Problem #2: Solve the IVP: $\displaystyle y"-y = t, y(0) = 0, y'(0) = 0$

Problem 3: For what values of w do solutions of $\displaystyle y"+9y = cos (wt)$ become unbounded as t goes to infinity. For what values of w are solutions periodic with a period of 3?

Problem #4: find the inverse Laplace Transform of the function $\displaystyle e^{-10_s}\frac{2}{s^{2}+4}$

thanks very much

Jen
• Mar 18th 2006, 11:02 PM
CaptainBlack
Quote:

Originally Posted by jenjen
Hi, I am studying for my final on Monday, 20th and I came across 3 questions on the 2nd order of differential equations that I couldn’t do. :( Please help me.

Problem #1: For what values of c do solutions y”+cy’+y = 0 have infinitely many zeros and tend to zero as t goes to infinity (here we are considering y as a function of t).

The solutions of:

$\displaystyle y''+cy'+y=0$

are all of the form:

$\displaystyle Ae^{k_1t}+Be^{k_2t}$

where $\displaystyle A$ and $\displaystyle B$ are arbitrary constants and
$\displaystyle e^{k_it},\ i=1,2$ are also solutions of the DE.

Now lets try $\displaystyle y(t)=e^{kt}$ as a trial solution of the equation:

$\displaystyle k^2y(t)+cky(t)+y(t)=0$

So:

$\displaystyle k^2+ck+1=0$

and hence:

$\displaystyle k=\frac{-c\pm\sqrt{c^2-4}}{2}$

Now if the $\displaystyle k$'s are real the solutions have at most a finite number
of zeros, so for there to be an infinite number of zeros of $\displaystyle y$
we need the $\displaystyle k$'s to be complex so then we must have $\displaystyle c<2$ (Note
this does not guarantee that the solution has any zeros for real $\displaystyle t$,
but solutions which are real will have an infinite number of zeros).

Also for $\displaystyle y(t)$ to die away as $\displaystyle t \to \infty$ we need
that $\displaystyle \Re k<0$, so $\displaystyle c>0$. This is because:

$\displaystyle y(t)=e^{(a+ib)t}$

is oscillatory with increasing amplitude if $\displaystyle a>0$, of constant amplitude
if $\displaystyle a=0$ and of decaying amplitude if $\displaystyle a<0$.

Hence for $\displaystyle y(t)$ to have an infinite number of zeros and to die
away as $\displaystyle t \to \infty$ we need $\displaystyle 0<c<2$

RonL
• Mar 18th 2006, 11:19 PM
jenjen
Quote:

Originally Posted by CaptainBlack
where $\displaystyle A$ and $\displaystyle B$ are arbitrary constants and
$\displaystyle e^{k_it},\ i=1,2$ are also solutions of the DE.

why did you choose 1, 2 for i? but not other numbers?
• Mar 18th 2006, 11:34 PM
CaptainBlack
Quote:

Originally Posted by jenjen
Problem #2: Solve the IVP: y”-y = t, y(0) = 0, y’(0) = 0

The general solution of

$\displaystyle y''-y=t$,

may be written as $\displaystyle h(t)+p(t)$, where $\displaystyle h(t)$ is the
general solution of $\displaystyle y''-y=0$, and $\displaystyle p(t)$ is a particular
solution (that is any solution of) of $\displaystyle y''-y=t$.

Now we know that the general solution of $\displaystyle y''-y=0$ is:

$\displaystyle h(t)=Ae^t+Be^{-t}$

Also it is clear that $\displaystyle p(t)=-t$ is a particular solution of $\displaystyle y''-y=t$,
so the general solution is:

$\displaystyle y(t)=Ae^t+Be^{-t} -t$.

Now we apply the initial conditions and these give:

$\displaystyle A+B=0$$\displaystyle A-B-1=0$,

so $\displaystyle A=1/2$ and $\displaystyle B=-1/2$. Hence:

$\displaystyle y(t)=\frac{1}{2}e^t-\frac{1}{2}e^{-t} -t$,

is the solution to the IVP.

RonL
• Mar 18th 2006, 11:40 PM
CaptainBlack
Quote:

Originally Posted by jenjen
why did you choose 1, 2 for i? but not other numbers?

because there are only two solutions to the quadratic that you get by
assuming a exponential solution and $\displaystyle k_1$ and $\displaystyle k_2$
are these two solutions, thats all there is to it.

The subscript is just an index indicating which solution for $\displaystyle k$

RonL
• Mar 19th 2006, 11:00 AM
jenjen
thanks to you CaptainBlack you are really helpful :) but I'm still studying hard for my final and I'm still confused on that one and none of my friends are able to do that so...

Problem #4: find the inverse Laplace Transform of the function $\displaystyle e^{-10_s}\frac{2}{s^{2}+4}$
• Mar 19th 2006, 11:35 AM
CaptainBlack
Quote:

Originally Posted by jenjen
thanks to you CaptainBlack you are really helpful :) but I'm still studying hard for my final and I'm still confused on that one and none of my friends are able to do that so...

Problem #4: find the inverse Laplace Transform of the function $\displaystyle e^{-10_s}\frac{2}{s^{2}+4}$

There is some confusion here, I'm not clear on what the $\displaystyle e^{-10_s}$
is supposed to denote, I guess it is supposed to be $\displaystyle e^{-10s}$,
is that right?

RonL
• Mar 19th 2006, 11:47 AM
jenjen
Quote:

Originally Posted by CaptainBlack
There is some confusion here, I'm not clear on what the $\displaystyle e^{-10_s}$
is supposed to denote, I guess it is supposed to be $\displaystyle e^{-10s}$,
is that right?

RonL

yeah you are right. sorry.
• Mar 19th 2006, 11:50 AM
CaptainBlack
Quote:

Originally Posted by jenjen
Problem #4: find the inverse Laplace Transform of the function $\displaystyle e^{-10s}\frac{2}{s^{2}+4}$

First thing we need is the convolution theorem for Laplace transforms:

$\displaystyle \mathcal{L}^{-1} f(s)g(s) = \int_0^t F(u)G(t-u)\ du$

Then we need the following:

$\displaystyle \mathcal{L}^{-1} e^{-as}=\delta (t-a)$,

and:

$\displaystyle \mathcal{L}^{-1} \frac{1}{s^2+a^2}=\frac{\sin(at)}{a}$.

So putting this all together we have:

$\displaystyle \mathcal{L}^{-1} e^{-10s}\frac{2}{s^2+4}=\int_0^t \delta (u-10) \sin(2(t-u)) \ du$,

which from the definition of the $\displaystyle \delta$ functional gives:

$\displaystyle \mathcal{L}^{-1} e^{-10s}\frac{2}{s^2+4}=\sin(2(t-10))\ \ t>10$
$\displaystyle =0\ \ t<10$

This also follows directly form the translation theorem.

RonL
• Mar 19th 2006, 12:03 PM
jenjen
Quote:

Originally Posted by CaptainBlack

$\displaystyle \mathcal{L}^{-1} \frac{1}{s^2+a^2}=\frac{\sin(at)}{a}$
RonL

why sin(at) and not cos(at) and where does the denominator (a) come from? my book formula states cos(at) without the denominator
• Mar 19th 2006, 12:07 PM
jenjen
nevermind, you are right!! :) sorry.
• Mar 19th 2006, 12:12 PM
jenjen
but I don't understand why you didn't have to convert $\displaystyle \mathcal{L}^{-1} e^{-10s}$ to 1/(s+10)
• Mar 19th 2006, 12:30 PM
jenjen
if we converted that to 1/(s+10), can we use partial fraction expansion to solve the problem? or we don't need partial fraction in this situation?
• Mar 19th 2006, 12:49 PM
CaptainBlack
Quote:

Originally Posted by jenjen
why sin(at) and not cos(at) and where does the denominator (a) come from? my book formula states cos(at) without the denominator

and my book has:

$\displaystyle \mathcal{L}^{-1} \frac{s}{s^2+a^2}=\cos(at)$

RonL
• Mar 19th 2006, 12:50 PM
CaptainBlack
Quote:

Originally Posted by jenjen
but I don't understand why you didn't have to convert $\displaystyle \mathcal{L}^{-1} e^{-10s}$ to 1/(s+10)

Not sure I understand this.

RonL
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