$\mathrm{restart}\;$

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{NumericalAnalysis}\right]\right)\:$

$\mathrm{DE1}≔\frac{\ⅆ}{\ⅆ t} y\left(t\right)\=y{\left(t\right)}^2-t$

$\mathrm{DE1}≔\frac{\ⅆ}{\ⅆt}y\left(t\right)={y\left(t\right)}^2-t$

$\mathrm{InitialValueProblem}\left(\mathrm{DE1}\, y\left(0\right)\=0.4\,t\=4\,\mathrm{method}\= \mathrm{adamsbashforth}\,\mathrm{output}\=\mathrm{solution}\right)\;$

$\mathrm{-3.933}$

$\mathrm{InitialValueProblem}\left(\mathrm{DE1}\, y\left(0\right)\=0.4\,t\=4\,\mathrm{method}\= \mathrm{adamsbashforth}\,\mathrm{output}\=\left[\mathrm{solution}\, \mathrm{Error}\right]\right)\;$

$\mathrm{-3.933},2.002$

$\mathrm{InitialValueProblem}\left(\mathrm{DE1}\, y\left(0\right)\=0.4\,t\=4\,\mathrm{method}\=\mathrm{rungekutta}\,\mathrm{submethod}\=\mathrm{rk4}\,\mathrm{output}\=\mathrm{plot}\right)\;$

$\mathrm{InitialValueProblem}\left(\mathrm{DE1}\, y\left(0\right)\=0.4\,t\=4\,\mathrm{method}\=\mathrm{rungekutta}\,\mathrm{submethod}\=\mathrm{rk4}\,\mathrm{comparewith}\=\left[\left[\mathrm{euler}\right]\, \left[\mathrm{adamsbashforth}\,\mathrm{step2}\right]\right]\, \mathrm{output}\=\mathrm{plot}\right)\;$

$\mathrm{InitialValueProblem}\left(\mathrm{DE1}\, y\left(0\right)\=0.4\,t\=4\,\mathrm{method}\=\mathrm{rungekutta}\,\mathrm{submethod}\=\mathrm{rk4}\,\mathrm{comparewith}\=\left[\left[\mathrm{euler}\right]\,\left[\mathrm{taylor}\,5\right]\right]\, \mathrm{output}\=\mathrm{information}\right)\;$

$\left[\begin{array}{cccccccc}t& \left[\mathrm{Maple\text{'}s\; numeric\; solution}\right]& \left[\mathrm{R-K\; 4th\; Ord.}\right]& \left[\mathrm{Error}\right]& \left[\mathrm{Euler}\right]& \left[\mathrm{Error}\right]& \left[\mathrm{5th-Ord.\; Taylor}\right]& \left[\mathrm{Error}\right]\\ 0.& 0.4& 0.4& 0.& 0.4& 0.& 0.4& 0.\\ 0.8000& 0.1842& 0.1907& 0.006540& 0.5280& 0.3438& 0.1779& 0.0063\\ 1.600& -0.6829& -0.6732& 0.009728& 0.1110& 0.7939& -0.6572& 0.0257\\ 2.400& -1.355& -1.311& 0.04412& -1.159& 0.1959& -1.381& 0.026\\ 3.200& -1.693& -1.618& 0.07503& -2.004& 0.3113& -1.692& 0.001\\ 4.& -1.931& -1.819& 0.1118& -1.351& 0.5804& -1.931& 0.\end{array}\right]$

$\mathrm{data1} ≔ \left[\left[0.422\,1.023\right]\,\left[3.533\,2.533\right]\,\left[4.022\,6.342\right]\,\left[5.002\,8.333\right]\,\left[6.003\,9.000\right]\right]$

$\mathrm{data1}≔\left[\left[0.422\,1.023\right]\,\left[3.533\,2.533\right]\,\left[4.022\,6.342\right]\,\left[5.002\,8.333\right]\,\left[6.003\,9.000\right]\right]$

$\mathrm{p1}≔\mathrm{PolynomialInterpolation}\left(\mathrm{data1}\, \mathrm{method}\=\mathrm{newton}\, \mathrm{function}\=\exp \left(x\right)\,\mathrm{independentvar}\=x\,\mathrm{errorboundvar}\=\mathrm{\ξ}\right)\:$

$\mathrm{Interpolant}\left(\mathrm{p1}\right)$

$0.8181719704+0.4853744777x+2.028886549\left(x-0.422\right)\left(x-3.533\right)-1.298772557\left(x-0.422\right)\left(x-3.533\right)\left(x-4.022\right)+0.4670460951\left(x-0.422\right)\left(x-3.533\right)\left(x-4.022\right)\left(x-5.002\right)$

$\mathrm{Function}\left(\mathrm{p1}\right)$

${\ⅇ}^x$

$\mathrm{RemainderTerm}\left(\mathrm{p1}\right)$

$\left(\frac{{\ⅇ}^{\mathrm{\xi }}\left(x-0.422\right)\left(x-3.533\right)\left(x-4.022\right)\left(x-5.002\right)\left(x-6.003\right)}{120}\right)\&where\left\{0.422\le \mathrm{\xi }\le 6.003\right\}$

$\mathrm{Draw}\left(\mathrm{p1}\,\mathrm{objects}\=\mathrm{BasisFunctions}\right)$

$\mathrm{p2}≔\mathrm{CubicSpline}\left(\mathrm{data1}\,\mathrm{independentvar}\=x\right)\:$

$\mathrm{Draw}\left(\mathrm{p2}\right)$

$\mathrm{M1}≔\mathrm{Matrix}\left(\left[\left[4.2\,2.2\,0.2\right]\,\left[1.0\,5.3\,2.1\right]\,\left[0.3\,3.2\,6.3\right]\right]\right)$

$\left[\begin{array}{ccc}4.2& 2.2& 0.2\\ 1.0& 5.3& 2.1\\ 0.3& 3.2& 6.3\end{array}\right]$

$\mathrm{P1}\,\mathrm{L1}\,\mathrm{U1}≔\mathrm{MatrixDecomposition}\left(\mathrm{M1}\, \mathrm{method}\=\mathrm{PLU}\right)$

$\mathrm{P1},\mathrm{L1},\mathrm{U1}≔\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0.2380952381& 1& 0\\ 0.07142857143& 0.6370887339& 1\end{array}\right],\left[\begin{array}{ccc}4.2& 2.2& 0.2\\ 0& 4.776190476& 2.052380952\\ 0& 0& 4.978165504\end{array}\right]$

$\mathrm{IsMatrixShape}\left(\mathrm{M1}\, \mathrm{strictlydiagonallydominant}\right)$

$\mathrm{true}$

$\mathrm{V1} ≔ \mathrm{Vector}\left(\left[1.5\,4.3\,3.2\right]\right)$

$\left[\begin{array}{c}1.5\\ 4.3\\ 3.2\end{array}\right]$

$\mathrm{LinearSolve}\left(\mathrm{M1}\,\mathrm{V1}\,\mathrm{method}\=\mathrm{SOR}\left(1.25\right)\right)$

$\left[\begin{array}{c}-0.05457376692\\ 0.7753875611\\ 0.1166895821\end{array}\right]$

$f≔x\to \sin \left(x\right)$

$f≔x\mapsto \sin \left(x\right)$

$\mathrm{int1}≔\sqrt{1\+{\left(\frac{\ⅆ}{\ⅆ x} f\left(x\right)\right)}^2}$

$\mathrm{int1}≔\sqrt{1+{\cos \left(x\right)}^2}$

$\mathrm{Quadrature}\left(\mathrm{int1}\, x\=1..14\,\mathrm{method}\=\mathrm{boole}\,\mathrm{output}\=\mathrm{value}\right)$

$15.74194117$

$\mathrm{Quadrature}\left(\mathrm{int1}\, x\=1..14\, \mathrm{method}\=\mathrm{boole}\,\mathrm{output}\=\mathrm{plot}\right)$

$\mathrm{AdaptiveQuadrature}\left(\mathrm{int1}\,x\=1..14\,\mathrm{method}\=\mathrm{newtoncotes}\left[4\right]\right)$

$15.74186755$

$\mathrm{AdaptiveQuadrature}\left(\mathrm{int1}\,x\=1..14\,\mathrm{method}\=\mathrm{newtoncotes}\left[4\right]\,\mathrm{output}\=\mathrm{information}\right)$

INTEGRAL: Int((1+cos(x)^2)^(1/2),x=1..14) = 15.7418533       
APPROXIMATION METHOD: Adaptive Bode's Rule
---------------------------------- INFORMATION TABLE ----------------------------------
    Approximate Value         Absolute Error         Relative Error
          15.7418676               1.426e-05         9.059e-05 %
---------------------------------- ITERATION HISTORY ---------------------------------------
   Interval           Status         Present Stack
      1..14            fail           EMPTY
      1..15/2          fail           [15/2, 14]
      1..17/4          fail           [[1], [17/4, 15/2]]
      1..21/8          PASS           [[2], [21/8, 17/4]]
   21/8..17/4          PASS           [[1], [17/4, 15/2]]
   17/4..15/2          fail           [15/2, 14]
   17/4..47/8          PASS           [[1], [47/8, 15/2]]
   47/8..15/2          PASS           [15/2, 14]
   15/2..14            fail           EMPTY
   15/2..43/4          fail           [43/4, 14]
   15/2..73/8          PASS           [[1], [73/8, 43/4]]
   73/8..43/4          PASS           [43/4, 14]
   43/4..14            fail           EMPTY
   43/4..99/8          PASS           [99/8, 14]
   99/8..14            PASS           EMPTY
--------------------------------------------------------------------------------------------
Number of Function Evaluations:     65

$f≔\ln \left(1-x\right)\+\cos \left(1-x\right)$

$f≔\ln \left(1-x\right)+\cos \left(-1+x\right)$

$\mathrm{Roots}\left(f\,x\=\left[0\,1\right]\,\mathrm{method}\=\mathrm{bisection}\right)$

$0.6022644043$

$\mathrm{Roots}\left(f\,x\=.4\,\mathrm{stoppingcriterion}\=\mathrm{absolute}\,\mathrm{method}\=\mathrm{newton}\,\mathrm{output}\=\mathrm{plot}\right)$

$\mathrm{IterativeFormulaTutor}\left(\right)$