[RESOLVED]Diff - differential
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12-06-2016, 12:26 PM
(This post was last modified: 12-06-2016 12:37 PM by Han.)
Post: #4
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RE: Diff - differential
Usually a "derivative" is involved when solving a differential equation. For example, \( y = G_0 \cos(x) + G_1 \sin(x) \) is a general solution to the equation \( \frac{d^2}{dx^2} y(x) + y(x) = 0 \). The equation \( \frac{d^2}{dx^2} y(x) + y(x) = 0 \) can also be written as \( y''(x) + y(x) = 0 \). So in this example, the differential equation \( y''(x) + y(x) = 0 \) involves the second derivative: \( y''(x) \). To solve the equation \( y''(x) + y(x) = 0 \) we use:
desolve(y''(x)+y(x) = 0, y) However, the equation you provided: \( z = 3x^2-xy+y^2\) does not involve any (partial) derivative or differential. Google translate: Por lo general, un "derivado" está involucrado cuando se resuelve una ecuación diferencial. Por ejemplo, \(y = G_0 \cos (x) + G_1 \sin (x) \) es una solución general a la ecuación \(\frac {d^2} {dx^2} y (x) + y(x) = 0 \). La ecuación \(\frac {d^2} {dx^2} y (x) + y(x) = 0 \) también se puede escribir como \(y '' (x) + y (x) = 0\). Así, en este ejemplo, la ecuación diferencial \(y '' (x) + y (x) = 0 \) implica la segunda derivada: \(y '' (x) \). Para resolver la ecuación \(y '' (x) + y (x) = 0 \) utilizamos: Desolve (y '' (x) + y (x) = 0, y) Sin embargo, la ecuación que proporcionó: \(z = 3x ^ 2-xy + y ^ 2 \) no implica ninguna derivada (parcial) o diferencial. Graph 3D | QPI | SolveSys |
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Messages In This Thread |
[RESOLVED]Diff - differential - jrozsas - 12-05-2016, 02:30 PM
RE: Diff - differential - Han - 12-05-2016, 09:24 PM
RE: Diff - differential - jrozsas - 12-06-2016, 12:03 PM
RE: Diff - differential - Han - 12-06-2016 12:26 PM
RE: Diff - differential - roadrunner - 12-06-2016, 12:27 PM
RE: Diff - differential - jrozsas - 12-06-2016, 01:41 PM
RE: Diff - differential - Han - 12-06-2016, 12:36 PM
RE: Diff - differential - jrozsas - 12-06-2016, 01:50 PM
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