Differential Equation Solutions - y'' = -y & y' = y

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Discussion Overview

The discussion revolves around the solutions to the differential equations y'' = -y and y' = y. Participants explore the nature of these solutions, questioning whether they are unique or if alternative forms exist. The scope includes theoretical aspects of differential equations and their general solutions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that the solution to y'' = -y is y = A*sin(x) + B*cos(x) for constants A and B, but question whether this is the only or most popular solution.
  • One participant claims that y = Ce^x is the only solution for the equation y' = y, while others do not explicitly confirm this.
  • Another participant introduces the idea that the solution y = A*sin(x) + B*cos(x) can also be expressed as y = A*sin(x + φ), where φ is a phase constant, suggesting this form is often used in physics.
  • There is a recognition that every differential equation of the form y^{(n)}(x)=f(x,y(x),...,y^{(n-1)}(x) has n independent solutions, implying that the general solutions for both equations are valid but may not be unique in form.

Areas of Agreement / Disagreement

Participants express varying degrees of certainty about the uniqueness of the solutions, with some asserting that there are alternative forms while others maintain a more definitive stance on the solutions presented. The discussion remains unresolved regarding whether the proposed forms are the only solutions.

Contextual Notes

Participants reference general solutions and alternative representations without resolving the implications of these forms or their equivalence in different contexts.

snipez90
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Hey guys, I was going through my differential equations packet and an example exercise illustrated that the differential equation y'' = -y has solution y = A*sin(x) + B*cos(x) for constants A and B. I understand why this is true but I was wondering if this was the only solution or at the very least if is most popular one. This also leads me to wonder if the diff eq y' = y has y = Ce^x as its only solution for a constant C.
 
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Yes that is the only solution.
Yes, Ce^x is the only solution of y' = y.
 
snipez90 said:
Hey guys, I was going through my differential equations packet and an example exercise illustrated that the differential equation y'' = -y has solution y = A*sin(x) + B*cos(x) for constants A and B. I understand why this is true but I was wondering if this was the only solution or at the very least if is most popular one. This also leads me to wonder if the diff eq y' = y has y = Ce^x as its only solution for a constant C.

Every differential equation

[tex]y^{(n)}(x)=f\left(x,y(x),\dots,y^{(n-1)}(x)\right)[/tex]

has n independent solutions. Thus

[tex]y''(x)+y(x)=0 \Rightarrow y(x)=A\,\cos x+B\,\sin x \rightarrow \text{General Solution}[/tex]

[tex]y'(x)-y(x)=0 \Rightarrow y(x)=C\,e^x \rightarrow \text{General Solution}[/tex]
 
snipez90 said:
Hey guys, I was going through my differential equations packet and an example exercise illustrated that the differential equation y'' = -y has solution y = A*sin(x) + B*cos(x) for constants A and B. I understand why this is true but I was wondering if this was the only solution or at the very least if is most popular one. This also leads me to wonder if the diff eq y' = y has y = Ce^x as its only solution for a constant C.


there is another popular but completely equivalent way to write the solution: [tex]y= A sin (x + \phi)[/tex] where phi is the phase constant. (Of course, you may as well use cos instead of sine). That's a form often used in phyiscs because it shows clearly that the solution is a simple sinusoidal with phase constant and amplitude determined by the initial conditions.
 
whoa, thanks for the replies. the solution to y'' = -y seems much less obvious than y' = y. very nice result though.
 

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