Math Help - Understanding the Definition of a "Solution" of an ODE

1. Understanding the Definition of a "Solution" of an ODE

Hi.

I'm having a little trouble getting a clear understanding of what is meant by the "interval of definition". I have Zill's A First Course in Differential Equations with Modeling Applications, and here is the definition given of a solution:

Definition 1.1.2: Solution of an ODE

Any function $\phi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n^{th}$ order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.

Now, I just don't understand what interval is meant by $I$. Correct me if my understanding is wrong. $I$ is the intersection of the domain of $\phi$ and the domain of the DE. If this isn't quite right, please explain it to me.

P.S. Some examples of the form "Given that $\phi(x)$ is a solution to the first order (or whatever order) DE $F(x,y,...,y^{n-1},y^n)=0$, state the interval of definition." would be great.

Thank You,
VN19.

2. Originally Posted by VonNemo19
Hi.

I'm having a little trouble getting a clear understanding of what is meant by the "interval of definition". I have Zill's A First Course in Differential Equations with Modeling Applications, and here is the definition given of a solution:

Definition 1.1.2: Solution of an ODE

Any function $\phi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n^{th}$ order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.

Now, I just don't understand what interval is meant by $I$. Correct me if my understanding is wrong. $I$ is the intersection of the domain of $\phi$ and the domain of the DE. If this isn't quite right, please explain it to me.

P.S. Some examples of the form "Given that $\phi(x)$ is a solution to the first order (or whatever order) DE $F(x,y,...,y^{n-1},y^n)=0$, state the interval of definition." would be great.

Thank You,
VN19.
What page? I have this book.

3. Originally Posted by dwsmith
What page? I have this book.
pg. 5 in the ninth edition.

4. This is best seen looking at an example. Work out Example 1 b on pg 5 as well.

$\displaystyle\frac{dy}{dx}=xy^{1/2} \ \ \ y=\frac{1}{16}x^4$

$\displaystyle\frac{dy}{dx}=\frac{1}{4}x^3$

Plug in.

$\displaystyle\frac{x^3}{4}=x\left(\frac{x^4}{16}\r ight)^2\Rightarrow\frac{x^3}{4}=x\left(\frac {x^2}{4}\right)\Rightarrow\frac{x^3}{4}=\frac{x^3} {4}$

Whenever you solve a DE, you can take your y you obtain and differentiate the amount of times need to check your solution.

Example

Say you have
$y''-5y'+2y+10=0$

When you find y, you can take the second derivative, minus 5 times the first, plus 2 times y, plus 10. The nth derivatives plugged into the DE should equal the RHS.

5. Originally Posted by dwsmith
.... The nth derivatives plugged into the DE should equal the RHS.
I don't think you quite understand what it is that I am asking. Thanks for trying anyway. I know how to plug in a solution and see if the equation reduces to an identity

I'm looking for an understanding of how to find the interval of definition of a solution.

6. Originally Posted by VonNemo19
I don't think you quite understand what it is that I am asking. Thanks for trying anyway.

I'm looking for an understanding of how to find the interval of definition. I know how to plug in a solution and see if the equation reduces to an identity
Domain of the solution.

7. I don't know what you mean by the "domain of the differential equation". I presume you mean the domain of the function F(x,t) where the differential equation is x'= F(x, t). No, the "domain of definition" is not the intersection of the domain of F with the domain of $\phi$. For one thing, F(x, t) is a function of two variables so its domain is in $R^2$ while the domain of the function $\phi$ is in $R$. For another, it makes no sense to say the domain of definition is domain of $\phi$ since $\phi$ is the solution. Of course, its domain is the "domain of definition" of the solution.

What is true of the "domain of definition", and probably what you intended to say, is that it must be a subset of projection of the domain of F(x,t) onto x-coordinate. Exactly what subset can be very difficult to determine.

8. Originally Posted by HallsofIvy
...
OK. So, let me try something else because I'm still confused about the INTERVAL OF DEFINITION; AKA "interval of existence"; AKA "interval of validity"; AKA "domain of the solution" <-----This one's in the book too, so it is a valid way to describe the idea that I'm trying to understand.

So, I'm gonna give a DE along with its solution and if someone could give the interval of definition and explain their reasoning as to how they determined that this interval was in fact the interval of defintion, I would really appreciate it.

Here we go:

Given that $x^2+y^2=25$ is a solution of the DE $\frac{dy}{dx}=-\frac{x}{y}$. State the interval of definition. Explain your reasoning.

BTW, thanks for the clarification regarding the multivariable DE and it's single variable solution HallsOfIvy

9. Originally Posted by VonNemo19
OK. So, let me try something else because I'm still confused about the INTERVAL OF DEFINITION; AKA "interval of existence"; AKA "interval of validity"; AKA "domain of the solution" <-----This one's in the book too, so it is a valid way to describe the idea that I'm trying to understand.

So, I'm gonna give a DE along with its solution and if someone could give the interval of definition and explain their reasoning as to how they determined that this interval was in fact the interval of defintion, I would really appreciate it.

Here we go:

Given that $x^2+y^2=25$ is a solution of the DE $\frac{dy}{dx}=-\frac{x}{y}$. State the interval of definition. Explain your reasoning.

BTW, thanks for the clarification regarding the multivariable DE and it's single variable solution HallsOfIvy
$x\in[-5,5] \ \ \ y\in[-5,5]$

10. Originally Posted by dwsmith
$x\in[-5,5] \ \ \ y\in[-5,5]$
So, basically, the interval of definition is simply the domain of the solution? Is this ALWAYS the case?

11. Yes, one of the its names is "domain of solution"