# Initial Value Problem

1. Jan 17, 2008

### Clutch Cargo

1. The problem statement, all variables and given/known data

I'm having problems with this IVP

dy/dx=y^4-x^4 and y(0)=7 I know the answer is yes but I just don't see how to get it.

2. Relevant equations

3. The attempt at a solution

2. Jan 17, 2008

### Gib Z

:( ? What do you mean, you know the answer is yes? Can you please show any work at all? Perhaps even state what specific type of differential equation this is, and tell us what other types you've ran into before. We can't help you unless we know how much you know.

3. Jan 17, 2008

### Clutch Cargo

I know that y(0)=7 is a solution to this IVP because I looked in the back of the book. I don't see how to prove it. My textbook only has one example IVP and it is nothing like this.

4. Jan 17, 2008

### Mathdope

y(0) = 7 isn't the answer, it's part of the problem.

5. Jan 17, 2008

### Clutch Cargo

Near as I can tell y'=dy/dx=y^4=x^4 is contiuous for y=0 and y=7

and f(x)=integral(y')=xy^4 -(x^5/5) is continuous for all x so IVP has a unique solution at y(0)=7.

Does this sound right to anyone?

6. Jan 17, 2008

### Dick

In common with other people, I don't know what you are talking about. If that's a differential equation with an initial value shouldn't you just try to solve it? In a correct way. I.e. not the way you did it. y is a function of x. You can't integrate y(x)^4 and get x*y(x)^4.

7. Jan 18, 2008

### Clutch Cargo

Thanks Dick. I feel much better that I am not the only one who doesn't understand what is going on. According to the textbook given:

dy/dx=f(x,y) and y(xo)=yo if f and df/dy are continuous then the problem has a unique solution. The initial value of y(0)=7 that is given is supposedly the unique solution. However if I plug dy/dx=y^4-x^4 into Lars Fredrickson's solvd() program the calculator dutifly spits out ln1 as the solution to the DE. I would look in my instructors solutions manual but it only has even numbered problems shown. The book has odd numbered answers in the back but no work shown. The problem I seem to be having is if the solution to the DE is ln1 what the **** is y(0)=7 supposed to mean?

8. Jan 18, 2008

### Mathdope

You seem to be misunderstanding things here, or something in your text is way out of whack. Either way:

What is the exact statement of the problem? It would appear to most of us here that you are given the equation dy/dx=y^4-x^4 with the initial condition y(0)=7. To be precise, they are asking you to solve for y as a function of x (i.e. y = f(x) ) where your answer should also satisfy f(0) = 7. The correct answer is not "y(0) = 7", it's an essential element in finding the answer.

What section of the text does this problem follow? You are likely supposed to use the technique that is described in it (unless it covers theory such as the existence and uniqueness theorem that you ref'd above).

9. Jan 18, 2008

### Clutch Cargo

The existence and uniqueness theorem is exactly what this problem is about. As I have stated the book only gives one example and it is nothing like this problem.

I have however looked at the solutions manual for other similar problems and I find in each case the answer book is only concerned with whether f(x,y) and f'(x,y) are continuous and it totally ignored the initial value given (in this case the y(0)=7)

I am wondering if the textbook is intentionally trying to make this confusing or what.

10. Jan 18, 2008

### HallsofIvy

Staff Emeritus
It's only in your fifth post that you tell us that this problem is about "the existence and uniqueness theorem" and you think your textbook is being confusing?

Now, as you were asked before, what is the exact statement of the problem? What are you trying to do?

Last edited: Jan 18, 2008
11. Jan 18, 2008

### Clutch Cargo

It says:

In problem 23-28 determine whether Theorem 1 (existence and uniqueness theorem) implies that the given initial value problem has a unique solution.

23. dy/dx=y^4-x^4 y(0)=7

12. Jan 18, 2008

### HallsofIvy

Staff Emeritus
Good! That says absolutely nothing about solving the equation!

Now what does "Theorem 1" say? In particular, do the hypotheses apply in this case?

13. Jan 18, 2008

### Clutch Cargo

It says that given the initial value of a problem:

dy/dx=f(x,y) y(xo)=yo

assume that f and df/dy are continuous fuctions in a rectangle

R={(x,y):a<x<b, c<y<d}

that contains the point (xo,yo). Then the initial value problem has a unique solution #(x)
in some interval xo-epsilon<x<xo+epsilon where epsilon is a positive number.

I know that y'=y^4-x^4 is continuous dy'/dxy=0 which is contiuous and integral(dy,dx)=xy^4-X^5/5 which is continuous so it seems that the theorem holds true.

Last edited: Jan 18, 2008
14. Jan 18, 2008

### Mathdope

The theorem applies to this particular problem. So...
Correct. This takes care of what the theorem refers to as f(x,y).
This doesn't make sense, please phrase it more carefully. To ensure that the IVP has a unique solution there are two functions which need to be verified as continuous. You've stated above that f(x,y) is continuous...what else do you need to check?

[By the way, I think the theorem actually wants you to check the partial of f with respect to y]

Last edited: Jan 18, 2008