1. ## Diff EQ

Two questions; the first is more of a check with some minor help:

1.) From calculus, what function(s) do you know where its 2nd deriv. is itself? What function(s) do you know where its 2nd deriv. is the negative of itself? Write each of the answers from above in the form of a diff. eq. with a solution.

Well, I know if f(x) = e^(x), then f''(x) = e^(x). And, I know if f(x) = cos(x), then f''(x) = -cos(x).

I'm not really sure of any other functions for those, unless you use f(x) = sin(x) for the second part, then f''(x) = -sin(x).

I guess you could use f(x) = e^(-x); then f''(x) = e^(-x)

I'm not sure how to answer these in the form of a diff. eq. with a solution unless it looks something like:

d^2y/dx^2 = -x where x = e^(x)... x = cos(x)... etc...? Uh, something like that? I'm not sure...

2.) I'm not sure where to begin on this one:

Find the region of the ty-plane where the diff. equation has a unique solution whose graph passes through a point (t_0, y_0) in that region.

(1 + y^3)y' = t^2

2. Originally Posted by Ideasman
Two questions; the first is more of a check with some minor help:

1.) From calculus, what function(s) do you know where its 2nd deriv. is itself? What function(s) do you know where its 2nd deriv. is the negative of itself? Write each of the answers from above in the form of a diff. eq. with a solution.

Well, I know if f(x) = e^(x), then f''(x) = e^(x). And, I know if f(x) = cos(x), then f''(x) = -cos(x).

I'm not really sure of any other functions for those, unless you use f(x) = sin(x) for the second part, then f''(x) = -sin(x).

I guess you could use f(x) = e^(-x); then f''(x) = e^(-x)

I'm not sure how to answer these in the form of a diff. eq. with a solution unless it looks something like:

d^2y/dx^2 = -x where x = e^(x)... x = cos(x)... etc...? Uh, something like that? I'm not sure...
-e^(-x) works for the second also.

to express this concept as a differential equation we simply do what the description says:

the function whose second derivative is itself:

let the function be $y$, then we are looking for solutions to

$y'' = y$

or

$y'' - y = 0$

the function whose second derivative is the negative of itself:

let the function be $y$, then we are looking for solutions to
$y'' = -y$

or

$y'' + y = 0$

3. Originally Posted by Ideasman
1.) From calculus, what function(s) do you know where its 2nd deriv. is itself? What function(s) do you know where its 2nd deriv. is the negative of itself? Write each of the answers from above in the form of a diff. eq. with a solution.
Actually, as it turns out you can use any number of them:
$y = Ae^{x} + Be^{-x}$
is the general solution to
$y^{\prime \prime} = y$

-Dan

4. Originally Posted by Ideasman
2.) I'm not sure where to begin on this one:

Find the region of the ty-plane where the diff. equation has a unique solution whose graph passes through a point (t_0, y_0) in that region.

(1 + y^3)y' = t^2
Taking all of the "clutter" from the problem you are looking for a solution of
$(1 + y^3)y^{\prime} = t^2$
that passes through the point $(t_0, y_0)$.

There are a few ways to do this one, but the one that leaps to mind from the form of the equation is the fact that the differential equation is separable:
$(1 + y^3) \frac{dy}{dt} = t^2$

$(1 + y^3) dy = t^2~dt$

$\int (1 + y^3) dy = \int t^2~dt$

$y + \frac{1}{4}y^4 = \frac{1}{3}t^3 + C$

We know this passes through $(t_0, y_0)$ so:
$y_0 + \frac{1}{4}y_0^4 = \frac{1}{3}t_0^3 + C$

Thus
$C = y_0 + \frac{1}{4}y_0^4 - \frac{1}{3}t_0^3$

So the (implicit) solution is
$y + \frac{1}{4}y^4 = \frac{1}{3}t^3 + \left ( y_0 + \frac{1}{4}y_0^4 - \frac{1}{3}t_0^3 \right )$

-Dan

5. Originally Posted by topsquark
Taking all of the "clutter" from the problem you are looking for a solution of
$(1 + y^3)y^{\prime} = t^2$
that passes through the point $(t_0, y_0)$.

There are a few ways to do this one, but the one that leaps to mind from the form of the equation is the fact that the differential equation is separable:
$(1 + y^3) \frac{dy}{dt} = t^2$

$(1 + y^3) dy = t^2~dt$

$\int (1 + y^3) dy = \int t^2~dt$

$y + \frac{1}{4}y^4 = \frac{1}{3}t^3 + C$

We know this passes through $(t_0, y_0)$ so:
$y_0 + \frac{1}{4}y_0^4 = \frac{1}{3}t_0^3 + C$

Thus
$C = y_0 + \frac{1}{4}y_0^4 - \frac{1}{3}t_0^3$

So the (implicit) solution is
$y + \frac{1}{4}y^4 = \frac{1}{3}t^3 + \left ( y_0 + \frac{1}{4}y_0^4 - \frac{1}{3}t_0^3 \right )$

-Dan
Thanks topsquark, but the question asks me to find a region where the diff. equation has a unique solution whose graph passes through a point (t_0, y_0) in that region. I don't think it wants the implicit solution.

6. Originally Posted by Ideasman
Thanks topsquark, but the question asks me to find a region where the diff. equation has a unique solution whose graph passes through a point (t_0, y_0) in that region. I don't think it wants the implicit solution.
I can't help you with all of it, but I can possibly give you a starting point:

We have existence because we were able to derive a solution. So there has to be some region where this solution exists, even if it is only the point $(t_0, y_0)$. So now we ask the question is there any part of the ty plane that the solution
$y + \frac{1}{4}y^4 = \frac{1}{3}t^3 + \left ( y_0 + \frac{1}{4}y_0^4 - \frac{1}{3}t_0^3 \right )$
is invalid? One way to explore this is to graph it. (See below where I have arbitrarily set $(t_0, y_0) = (0, 1)$.)

First note that this is not a function. Also note that there is a domain of t where the solution does not exist.