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Y'' = -y

  1. Jan 13, 2008 #1
    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.
  2. jcsd
  3. Jan 13, 2008 #2


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    Yes that is the only solution.
    Yes, Ce^x is the only solution of y' = y.
  4. Jan 13, 2008 #3
    Every differential equation


    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]
  5. Jan 13, 2008 #4


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    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.
  6. Jan 13, 2008 #5
    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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