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Wolfram mathematica . . .How to solve typical mathematics textbook problems on the most advanced computer algebra system in the world. Roger Wells July 2023{“Kernel” -> {“Version” -> “11.2.0 for Microsoft Windows (64-bit) (September 11, 2017)”, “ReleaseID” -> “11.2.0.0 (5822651, 2017091101)”, “LicenseID” -> “L3991-4623”, “PatchLevel” -> “0”, “ActivationKey” -> “3991-4623-49AQHW”, “ActivationGroupID” -> “L3991-4623”, “MachineID” -> “6202-12511-69189”, “UserName” -> “Owner”, “MachineName” -> “tyrannasaurus-r”, “MachineDomains” -> {}, “LicenseServer” -> “tyrannasaurus-r”, “MaxLicenseProcesses” -> 2, “LicenseExpirationDate” -> {Infinity, 0, 0, 0, 0, 0}, “MachineType” -> “PC”, “OperatingSystem” -> “Windows”, “ProcessorType” -> “x86-64”, “Language” -> “English”, “CharacterEncoding” -> “WindowsANSI”, “SystemCharacterEncoding” -> “WindowsANSI”, “TimeZone” -> -6., “CreationDate” -> DateObject[{2017, 9, 11, 10, 31, 25.}, “Instant”, “Gregorian”, -6.], “InstallationDirectory” -> “C:\\Program Files\\Wolfram Research\\Mathematica\\11.2”,Machine Epsilon 2022045 * 10^-16 Machine precision 159546 Machine Number Limits { 2.225.7 * 10^ -308, 1,79769 *10^ 308} Number Limits {6,22968849675322 *

-1355718576299610

, 16052167619336622 *

1355718576299609

} This is an active computable document created on software owned by Roger Wells January 2018 . Wolfram Mathematica ID: rogerwells2017@hotmail.com. PTC Mathcad username: iornfist666@yahoo.com Password: Javadog License Number 3991-4623 Password: 1javadog Mathcad serial # EN901002DS2411A https://user.wolfram.com/portal/ProductRegistration?execution=e1s Linkdln: fgmju@yahoo.com 1javadog!Quadratic EquationSolve for X

a(x)

+ b(x) +c = 0

Solve[ax^2+bx+c0,x]

x

-b-

-4ac

,x

-b+

-4ac



MatrixFormx

-b-

-4ac

,x

-b+

-4ac



x

-b-

-4ac

x

-b+

-4ac

Solve[ax^2+bx+c0,b]

b

-c-a



Solve[ax^2+bx+c0,c]

{{c-bx-a

}}

Matrix Algebra Parenthesis to open Matrix ()control + enter = Rowcontrol + comma = Column

			
			
			
			

The input (above) is Matrix Form (obviously). The software considers this to be a Graphic (picture)The output (below) is a List of Lists. This is available to be used as input for computation

{{,,,},{,,,},{,,,},{,,,}}

a=4;b=3;

c=3;

Assign these variables, then use them in a Matrix

d=2;

a	b
c	d

;

Inv=Inverse[{{a,b},{c,d}}]

{{-2,3},{3,-4}}

MatrixForm

-bc+ad

,-

-bc+ad





-2	3
3	-4



Differential CalculusDetermine the general solution.

2x+y

y-x

DSolvey'[x]

2x+y[x]

y[x]-x

,y[x],xyoumustcompletethearguments...,Y[x],XandclosetheBracet

Note the output is again a List with two arguments {{ x→ neg} , {x→pos}}

y[x]x-

2C[1]



,y[x]x+

2C[1]





In the Wolfram language



[ X ]

[ X ] Y = Y [ x ] ( keeping Y as a function of X )Determine the second order general solution.

- 2

- 8y = 0

= - 2

- 8y = 0

DSolve[y''[x]-2y'[x]-8y[x]==0,y[x],x] remembertousedoubleequalsign

{{y[x]

-2x



C[1]+



C[2]}}

Specific solutions and boundary conditions . . .DSolve[ { x’ [ t ] = = [ t ], x [ 0 ] = = }, x [ t ] , t ] Elementary displacement function f (x.) initial value zero

{{x[t]



}}

Determine the general solution of the differential equation

= k ( y - T )Assume the initial conditions Y(0) = T1A typical population growth model where T and T1 are (time ( 0 ) and time (1) ) “initial value and condition”Wolfram mathematica . . . DSolve [ { y ’ [ x ] = = k ( y [ x ] - T ) , y [ 0 ] = = T1 } , y [ x ], x ]DSolve [ { y’ [ x ]  k ( y [ x ] - T ) , y [ 0 ]  T1 } , y [ x ] , x ] note the space between k * (the constant) and k * ( y [ x ] - T ) multiplication implied Mathematica returns the output

{{y[x]T-



T+



T1}}

NDSolve[{y’[x] == Sin[y[x]], y[0] == 1}, y, {x, 0, 4}] With NDSolve the “interpolating” function appears.

yInterpolatingFunction

Domain: {{0.,4.}}

Output: scalar

Thesewindowsappearintheanswer.Theyarereadyforcalculationsonthenextline.

The windows can also be opened and contain a wealth of information and algorithms pertaining to the differential equation. % 1 [ [1, 1, 2] ] This now uses % which means input from last line (1).