Very basic Q about solns to y" = y

In summary, Wolfram and the Linear Algebra text both agree that the two possible solutions for the equation \frac{d^2y}{dx^2}=y are e^{x} and e^{-x}, or constant multiples of them. The added polynomial, y = e^x + x, is not a solution because when differentiated twice, it does not equal y. This is due to the fact that y=x is not a solution.
  • #1
kostoglotov
234
6
Wolfram and the Linear Algebra text I'm currently working on, give the two possible solutions of [itex]\frac{d^2y}{dx^2}=y[/itex] as being [itex]e^{x}[/itex] and [itex]e^{-x}[/itex], or rather, constant multiples of them.

Here wolfram agrees:

http://www.wolframalpha.com/input/?i=d^2y/dx^2=y

My question is, why isn't [itex]y = e^{x} + x[/itex] and [itex]y = e^{-x} + x[/itex] not solutions?

Or is an added polynomial in the solution just part of some nullspace of special solutions that we consider separately in general?

edit: wouldn't the added polynomial change the dimension of the solution space?
 
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  • #2
Why should y=e^x + x be a solution? y=x is not a solution.
 
  • #3
davidmoore63@y said:
Why should y=e^x + x be a solution? y=x is not a solution.

If you take y=e^x + x and diff it twice...damn it...ok, I see it now...this was dumb.
 

1. What is a solution to "y = y"?

A solution to this equation is any value of y that makes the statement true. This includes all real numbers, as well as infinity.

2. How many solutions does "y = y" have?

This equation has an infinite number of solutions, as any value of y will make the statement true.

3. Is "y = y" a linear equation?

Yes, "y = y" is a linear equation because it has a degree of 1 and can be written in the form y = mx + b, where m = 1 and b = 0.

4. Can "y = y" have any other form?

No, "y = y" is the simplest form of this equation and cannot be simplified any further.

5. How can I graph "y = y"?

Since this equation has an infinite number of solutions, the graph of "y = y" is a straight line that goes through all points on the coordinate plane.

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