Very basic Q about solns to y" = y

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SUMMARY

The discussion centers on the differential equation \(\frac{d^2y}{dx^2}=y\) and its solutions, specifically the exponential functions \(e^{x}\) and \(e^{-x}\). The user questions why functions like \(y = e^{x} + x\) and \(y = e^{-x} + x\) are not considered solutions. Upon further reflection, the user realizes that the added polynomial does not satisfy the original differential equation, confirming that only the exponential functions are valid solutions.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with exponential functions and their properties.
  • Basic knowledge of the concept of solution spaces in linear algebra.
  • Experience with differentiation and its application in solving differential equations.
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  • Study the general solutions of second-order linear differential equations.
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Students and professionals in mathematics, particularly those studying differential equations and linear algebra, as well as educators seeking to clarify concepts related to solution spaces and polynomial functions in differential equations.

kostoglotov
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Wolfram and the Linear Algebra text I'm currently working on, give the two possible solutions of \frac{d^2y}{dx^2}=y as being e^{x} and e^{-x}, 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 y = e^{x} + x and y = e^{-x} + x 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?
 
Physics news on Phys.org
Why should y=e^x + x be a solution? y=x is not a solution.
 
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.
 

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