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I How to solve the two following differential equations?

  1. Nov 21, 2016 #1
    (1) ##\frac{d^{2}y}{dx^{2}}=0##
    (2) ##\frac{d^{2}y}{dx^{2}}=k^{2}y##, where k is a real positive number.
     
  2. jcsd
  3. Nov 21, 2016 #2

    haushofer

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    If the second derivative of a function vanishes, it can at most be linear in x. Same reasoning for your second equation, although looking at the notation I suspect you mean the rhs to have y(x) instead of x. If that is so, you want to try an exponential function like e^[kx].
     
  4. Nov 21, 2016 #3
    Thank you for your answer. But for the first equation, is there any process of deduction?
     
  5. Nov 21, 2016 #4
    Thank you very much for pointing the spelling error in the second equation, I have modified it.
     
  6. Nov 21, 2016 #5
    I have searched google for "harmonic oscillator", however, it is a little different to the equation I give. For harmonic oscillator, the equation is like this, ##\frac{d^{2}y}{dx^{2}}=-k^{2}y##, where there is an additional negative sign.
     
  7. Nov 21, 2016 #6
    Now I know the solutions:
    (1)For $$\frac{d^{2}y}{dx^{2}}=0,$$ the solution is $$y=A+Bx.$$
    (2)For $$\frac{d^{2}y}{dx^{2}}=k^{2}y,$$ the solution is $$y=Ae^{-kx}+Be^{-kx}.$$
    But I still don't know how they are deduced.
     
  8. Nov 30, 2016 #7

    haruspex

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    1) is easy. Just integrate both sides twice, remembering to plug in a constant of integration.
    For 2), multiply both sides by dy/dx. Now you can integrate both sides. What do you get?
     
  9. Nov 30, 2016 #8

    LCKurtz

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    That is a constant coefficient linear DE of the form ##y'' + \lambda y = 0##. The standard technique for such an equation is to assume a solution of the form ##e^{rx}##. Plug that in and see what you get. You will want to consider three cases:$$
    \lambda = k^2 > 0,~\lambda = 0,~\lambda = -k^2 <0\text{.}$$
     
  10. Nov 30, 2016 #9

    haruspex

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    Sure, but I interpreted Tspirits' question as how to proceed if you have no idea what the answer is.
     
  11. Dec 1, 2016 #10

    lurflurf

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    $$\dfrac{d^2y}{{dx}^2}-k^2y=0$$
    we have haruspex's sugestion
    $$2\dfrac{d^2y}{{dx}^2}\dfrac{dy}{dx}-2k^2y\dfrac{dy}{dx}=0 \\
    \dfrac{d}{dx}\left(\dfrac{dy}{dx}+k y\right)\left(\dfrac{dy}{dx}-k y\right)=0$$
    we could factor the operator
    $$\left(\dfrac{d}{dx}+k \right)\left(\dfrac{d}{dx}-k \right)y=0$$
    we could solve for the inverse
    $$-\left(\dfrac{dx}{dy}\right)^{-3}\dfrac{d^2x}{{dy}^2}-k^2y=0$$
     
  12. Dec 1, 2016 #11

    haruspex

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    I was thinking more along the lines of
    y"=k2y
    2y'y"=2k2yy'
    y'2=k2y2+c2
    Then some suitable hyperbolic trig substitution.
     
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