1. Jun 3, 2013

### kevinnn

I'm only a first year calculus student so I don't feel too bad asking this probably really dumb question. I noticed in my textbook that the definition of a differential equation was an equation involving the derivatives of a function. Soooo wouldn't that mean that any equation where you can calculate an antiderivative is a differential equation? I don't know if there are any non differentiable equations but if there is not then cant every equation be considered a differential equation? I know differential equations are very important so time is required for me to learn about them and their importance. This is just a curious question I have now. Thanks.

2. Jun 3, 2013

### Simon Bridge

Solving a DE is not normally thought of as finding the antiderivative of it.

Consider the equation: y'' + ay' + by = c
What would you understand by "calculate an antiderivative of" that equation?

Also, you can calculate the antiderivative of many equations that are not differential equations.
eg.

find the antiderivative of y=x^2+ax+b

This is not an differential equations because it does not contain derivatives.

3. Jun 3, 2013

### HallsofIvy

Yes, finding an anti-derivative of, say, f(x) is precisely solving the differential equation dy/dx= f(x).

That would be the very simplest kind of differential equation and is given little time in texts on differential equations (since you would have learned them in Calculus).

(I almost agree with Simon Bridge although I am a little uncomfortable with his repeated reference to find anti-derivatives of equations rather than functions.)

4. Jun 3, 2013

### Simon Bridge

I know - I am using OPs words - notice how I am asking questions about what it would mean?

I'm expecting to modify the response depending on how OP answers.

5. Jun 3, 2013

### kevinnn

find the antiderivative of y=x^2+ax+b

This is not an differential equations because it does not contain derivatives.[/QUOTE]

That was just my question. What I was thinking is that it could be a differential equation because it contains x^2, which is the derivative of (1/3)x^3.

6. Jun 3, 2013

### Office_Shredder

Staff Emeritus
Your intuition is backwards. It is secretly a differential equation because they asked you to solve for z given dz/dz = x2 +ax+b

7. Jun 3, 2013

### Simon Bridge

That would be:
$1 = x^2 + ax + b$ then...

And your question has been answered - "no", merely having integrable functions in it is not sufficient to make it a differential equation.

y(x) could be the solution to the DE $y^\prime (x) = 2x+a : y(0)=b$.

A differential equation for y(x) would have to be in terms of y and derivatives of y (and an explicitly given function of x)

i.e. $$\sum_{i=0}^n A_i\frac{d^i}{dx^i}y(x)=f(x)$$ would be a differential equation[1] provided any $A_{i > 0} \neq 0$

This sum gives $y=x^2+ax+b$ if $f(x)=x^2+ax+b$ and $A_{i>0}=0$ so it isn't one.

You can look up the formal definitions for "differential equation" if you like.
The way you are thinking about it isn't going to be very useful to you.

8. Jun 6, 2013

### lurflurf

It depends what antiderivative is taken to mean
Many books would write
y'' + ay' + by = c
as
then write the anti (D^2+aD+b) of c

y=x^2+ax+b
is surely a differential equation you just confused yourself by writing y=Y'
Y'=x^2+ax+b
there I fixed it for you.

9. Jun 7, 2013

### Simon Bridge

@lurflurf: thank you - can you show me how these ideas relate to the question in post #1?

I was actually asking those question to explore OPs ideas about what the terms meant, not to clarify my own.
The fact there are many interpretations is the reason I wanted to do that.