- #1
Tspirit
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(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) ##\frac{d^{2}y}{dx^{2}}=k^{2}y##, where k is a real positive number.
Thank you for your answer. But for the first equation, is there any process of deduction?haushofer said: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].
Thank you very much for pointing the spelling error in the second equation, I have modified it.haushofer said: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].
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.haushofer said: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].
1) is easy. Just integrate both sides twice, remembering to plug in a constant of integration.Tspirit said: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.
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:$$Tspirit said:(2)For $$\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.
Sure, but I interpreted Tspirits' question as how to proceed if you have no idea what the answer is.LCKurtz said:The standard technique for such an equation is to assume a solution of the form
I was thinking more along the lines oflurflurf said:haruspex's sugestion
A differential equation is a mathematical equation that relates a function with its derivatives. It describes how a quantity changes over time or space, and is used to model many physical, biological, and economic phenomena.
The two types of differential equations are ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs involve a single independent variable, while PDEs involve multiple independent variables.
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