Texas Instruments voyage 200 User Manual - Page 806
if its absolute value is less than, This tolerance
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easier than determining a general solution, substituting initial values, solving for the arbitrary constant, and then substituting that value into the general solution. ë(2øsin(y)+yñ) 2 =ë(exì1)øeëxøsin(y) soln|x=0 and y=0 ¸ true initialCondition is an equation of the form: d(right(eq)ì left(eq),x)/ dependentVar (initialIndependentValue) = (d(left(eq)ì right(eq),y)) initialDependentValue ! impdif(eq,x,y) ¸ The initialIndependentValue and initialDependentValue Done can be variables such as x0 and y0 that have no ode|y'=impdif(soln,x,y) ¸ stored values. Implicit differentiation can help true verify implicit solutions. DelVar ode,soln ¸ Done deSolve(2ndOrderOde and initialCondition1 and initialCondition2, independentVar, dependentVar) ⇒ a particular solution Returns a particular solution that satisfies 2ndOrderOde and has a specified value of the dependent variable and its first derivative at one point. deSolve(y''=y^(ë 1/2) and y(0)=0 and y'(0)=0,t,y) ¸ 2øy 3 3/4 =t solve(ans(1),y) ¸ y= 22/3ø(3øt)4/3 4 and t,0 For initialCondition1, use the form: dependentVar (initialIndependentValue) = initialDependentValue For initialCondition2, use the form: dependentVar' (initialIndependentValue) = initial1stDerivativeValue deSolve(2ndOrderOde and boundaryCondition1 and boundaryCondition2, independentVar, dependentVar) ⇒ a particular solution Returns a particular solution that satisfies 2ndOrderOde and has specified values at two different points. deSolve(w''ì 2w'/x+(9+2/x^2)w= xù e^(x) and w(p/6)=0 and w(p/3)=0,x,w) ¸ e p 3øxøcos(3øx) w= 10 ì e p 6øxøsin(3øx) 10 + x⋅ex 10 det( ) MATH/Matrix menu det(squareMatrix[, tol]) ⇒ expression det([a,b;c,d]) ¸ aø d ì bø c Returns the determinant of squareMatrix. det([1,2;3,4]) ¸ ë2 Optionally, any matrix element is treated as zero if its absolute value is less than tol. This tolerance is used only if the matrix has floating-point entries and does not contain any symbolic variables that have not been assigned a value. Otherwise, tol is ignored. • If you use ¥ ¸ or set the mode to Exact/Approx=APPROXIMATE, computations are done using floating-point arithmetic. • If tol is omitted or not used, the default tolerance is calculated as: det(identity(3) ì xù [1,ë 2,3; ë 2,4,1;ë 6,ë 2,7]) ¸ ë (98ø xò ì 55ø xñ + 12ø x ì 1) [1E20,1;0,1]!mat1 det(mat1) ¸ det(mat1,.1) ¸ [01.E20 11] 0 1.E20 5Eë 14 ù max(dim(squareMatrix)) ù rowNorm(squareMatrix) 808 Appendix A: Functions and Instructions
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