# Thread: 2nd order of differential equations

1. ## 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

2. 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

3. 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?

4. 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

5. 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

6. 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 $\displaystyle e^{-10_s}\frac{2}{s^{2}+4}$

7. 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

8. 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.

9. 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

10. 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

11. nevermind, you are right!! sorry.

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

13. 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?

14. 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

15. 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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