simulink求解二階系統數值微分

對於 $[left{ egin{array}{l} mddot y + cdot y + ky = f(t)\ y(0) = {y_0}\ dot y(0) = {{dot y}_0} end{array} ight.]$ 形式的微積分可以採用如下形式的求解方式 $[egin{array}{l} y(k + 1) = a(Delta t)y(k) + b(Delta t)y(k - 1) + c(Delta t)f(k)\ a(Delta t) = 2frac{{2m - kDelta {t^2}}}{{2m + cDelta t}},b(Delta t) = frac{{cDelta t - 2m}}{{2m + cDelta t}},c(Delta t) = frac{{2Delta {t^2}}}{{2m + cDelta t}},k = 0,1,2,...,n end{array}]$

y(1)和y(-1)求解如下：

$[left{ egin{array}{l} y(1) = frac{1}{2}[frac{{Delta {t^2}}}{m}f(0) + (2Delta t - frac{{Delta {t^2}c}}{m})dot y(0) + (2 - frac{{Delta {t^2}k}}{m})y(0)]\ y( - 1) = frac{1}{2}[frac{{Delta {t^2}}}{m}f(0) - (2Delta t - frac{{Delta {t^2}c}}{m})dot y(0) + (2 - frac{{Delta {t^2}k}}{m})y(0)] end{array} ight.]$

特別的當時 $[y(1) = y( - 1) = frac{{Delta {t^2}}}{m}f(0)]$