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What's New in Education in Maple 2025?

 

Maple 2025 includes a number of improvements to support teaching and learning of mathematics and science.

 

Step-by-Step Solutions

Improved Ability to Generate Similar Problems

Try Another: A New Addition to Check My Work

Student:-ODEs

Radius of Convergence

Find a Formula for the nth Term of a Given Integer Sequence

Step-by-Step Solutions

Maple 2025 improves the existing suite of commands for showing step-by-step solutions to standard math problems.  It also adds one new method.

Our step-by-step solutions in Maple are highly regarded by our users and are widely used within Maple. These same solutions are also featured in Maple Calculator and Maple Learn, and we consistently receive valuable feedback on them across all our products. For Maple 2025, we’ve made exciting updates based on this input. We’ve expanded our Linear Algebra steps by adding steps for matrix transpose, and we’ve refined the heuristics for several integration methods, including integration by parts and change of variables, to make the steps clearer for students and more aligned with the approaches they will encounter. 

Step-by-Step Integration

Maple 2025 has improved heuristics for several classes of integration method, including integration by parts and change of variables.

 withStudent:-Calculus1:UnderstandInt, constant, `c*`, `+`:

Integration by Parts

Integration by parts is now utilized in more cases where it will lead to a simpler sequence of steps.

The following example is now done in far fewer steps, and also produces a much simpler result.  In previous versions the solution began with a u=ⅇ6⁢x substitution, leading to the result 5⁢sin⁡5⁢x⁢ⅇ6⁢x2+6⁢cos⁡5⁢x⁢ⅇ6⁢x2−5⁢sin⁡5⁢x+6⁢cos⁡5⁢x122⁢ⅇ6⁢x+C

ShowSolutionIntcos5⋅x⋅sinh6⋅ x, x

Integration Steps∫cos⁡5⁢x⁢sinh⁡6⁢xⅆx▫1. Apply integration by Parts◦Recall the definition of thePartsrule∫uⅆv=v⁢u−∫vⅆu◦First partu=cos⁡5⁢x◦Second partdv=sinh⁡6⁢x◦Differentiate first partdu=ⅆⅆxcos⁡5⁢xdu=−5⁢sin⁡5⁢x◦Integrate second partv=∫sinh⁡6⁢xⅆxv=cosh⁡6⁢x6∫cos⁡5⁢x⁢sinh⁡6⁢xⅆx=cos⁡5⁢x⁢cosh⁡6⁢x6−∫−5⁢cosh⁡6⁢x⁢sin⁡5⁢x6ⅆxThis gives:cos⁡5⁢x⁢cosh⁡6⁢x6+5⁢∫cosh⁡6⁢x⁢sin⁡5⁢xⅆx6▫2. Apply integration by Parts◦Recall the definition of thePartsrule∫uⅆv=v⁢u−∫vⅆu◦First partu=sin⁡5⁢x◦Second partdv=cosh⁡6⁢x◦Differentiate first partdu=ⅆⅆxsin⁡5⁢xdu=5⁢cos⁡5⁢x◦Integrate second partv=∫cosh⁡6⁢xⅆxv=sinh⁡6⁢x6∫cosh⁡6⁢x⁢sin⁡5⁢xⅆx=sin⁡5⁢x⁢sinh⁡6⁢x6−∫5⁢cos⁡5⁢x⁢sinh⁡6⁢x6ⅆxThis gives:cos⁡5⁢x⁢cosh⁡6⁢x6+5⁢sin⁡5⁢x⁢sinh⁡6⁢x6−5⁢∫cos⁡5⁢x⁢sinh⁡6⁢xⅆx66▫3. Solve the equation algebraically◦DefineF=∫cos⁡5⁢x⁢sinh⁡6⁢xⅆx◦Solve forFF=cos⁡5⁢x⁢cosh⁡6⁢x6+5⁢sin⁡5⁢x⁢sinh⁡6⁢x36−25⁢F36This gives:6⁢cos⁡5⁢x⁢cosh⁡6⁢x61+5⁢sin⁡5⁢x⁢sinh⁡6⁢x61•Add constant of integration6⁢cos⁡5⁢x⁢cosh⁡6⁢x61+5⁢sin⁡5⁢x⁢sinh⁡6⁢x61+C

(1)

 

Improved Change of Variables

An improved heuristic for finding successful candidates for a change of variable allows more examples to succeed, or to succeed in fewer steps.

This example was not able to be computed in previous versions:

ShowSolutionIntx+12/sinx2+x, x

Integration Steps∫x+12sin⁡x2+xⅆx▫1. Apply a change of variables to rewrite the integral in terms ofu◦Letubeu=−x2−x◦Differentiate both sidesdu=−2⁢x−1⁢dx◦Isolate equation fordxdx=−12⁢x+1⋅du◦Substitute the values for u and dx back into the original∫x+12sin⁡x2+xⅆx=∫csc⁡uⅆu2This gives:∫csc⁡uⅆu2▫2. Rewrite◦Equivalent expressioncsc⁡u=csc⁡u2−csc⁡u⁢cot⁡ucsc⁡u−cot⁡uThis gives:∫csc⁡u2−csc⁡u⁢cot⁡ucsc⁡u−cot⁡uⅆu2▫3. Apply a change of variables to rewrite the integral in terms ofv◦Letvbev=csc⁡u−cot⁡u◦Differentiate both sidesⅆvⅆu=ⅆⅆucsc⁡u−cot⁡u◦Evaluatedv=−csc⁡u⁢cot⁡u+1+cot⁡u2⁢du◦Substitute in the values ofuanddu∫csc⁡u2−csc⁡u⁢cot⁡ucsc⁡u−cot⁡uⅆu=∫1vⅆvThis gives:∫1vⅆv2▫4. Apply thereciprocalrule to the term∫1vⅆv◦Recall the definition of thereciprocalrule∫1vⅆv=ln⁡vWe can rewrite the integral as:12⋅ln⁡v▫5. Revert change of variable◦Variable we defined in step3v=csc⁡u−cot⁡uThis gives:12⋅ln⁡csc⁡u−cot⁡u▫6. Revert change of variable◦Variable we defined in step1u=−x2−xThis gives:12⋅ln⁡−csc⁡x2+x+cot⁡x2+x•Add constant of integrationln⁡−csc⁡x2+x+cot⁡x2+x2+C

(2)

This example was not able to be computed in previous versions:

ShowSolutionIntx4+7⋅sin15⋅x5+7⋅x,x 

Integration Steps∫x4+7⁢sin⁡15⁢x5+7⁢xⅆx▫1. Apply a change of variables to rewrite the integral in terms ofu◦Letubeu=15⁢x5+7⁢x◦Differentiate both sidesdu=x4+7⁢dx◦Isolate equation fordxdx=1x4+7⋅du◦Substitute the values for u and dx back into the original∫x4+7⁢sin⁡15⁢x5+7⁢xⅆx=∫sin⁡uⅆuThis gives:∫sin⁡uⅆu▫2. Evaluate the integral ofsin(u)◦Recall the definition of thesinrule∫sin⁡uⅆu=−cos⁡uThis gives:−cos⁡u▫3. Revert change of variable◦Variable we defined in step1u=15⁢x5+7⁢xThis gives:−1⋅cos⁡15⁢x5+7⁢x•Add constant of integration−cos⁡15⁢x5+7⁢x+C

(3)

The following example is now done in fewer steps.  The initual u-substitution now includes the -1 constant, thus avoiding the need for a second change of variables later in the solution.

ShowSolutionInt5⋅ x4−3⋅x2 −2⋅x⋅x5−x3−x2−1−1, x

Integration Steps∫5⁢x4−3⁢x2−2⁢xx5−x3−x2−1ⅆx▫1. Apply a change of variables to rewrite the integral in terms ofu◦Letubeu=x5−x3−x2−1◦Differentiate both sidesdu=5⁢x4−3⁢x2−2⁢x⁢dx◦Isolate equation fordxdx=1x⁢5⁢x3−3⁢x−2⋅du◦Substitute the values for u and dx back into the original∫5⁢x4−3⁢x2−2⁢xx5−x3−x2−1ⅆx=∫1uⅆuThis gives:∫1uⅆu▫2. Apply thereciprocalrule to the term∫1uⅆu◦Recall the definition of thereciprocalrule∫1uⅆu=ln⁡uWe can rewrite the integral as:ln⁡u▫3. Revert change of variable◦Variable we defined in step1u=x5−x3−x2−1This gives:ln⁡x5−x3−x2−1•Add constant of integrationln⁡x5−x3−x2−1+C

(4)

 

Integration Steps

Maple 2025 improves the steps for integration problems given by the ShowSolution command.  More detail is provided for change-of-variables integration steps.  The change of bounds step is documented when applying change-of-variable rule to a definite integral.

expr ≔ Int −x2+9, x=0..3 

expr≔∫03−x2+9ⅆx

(5)

ShowSolutionexpr

Integration Steps∫03−x2+9ⅆx▫1. Apply a change of variables to rewrite the integral in terms ofu◦Found patterna−b⁢x2; let:x=3⁢sin⁡u◦Differentiate both sidesⅆxⅆx=ⅆⅆu3⁢sin⁡u◦Evaluatedx=3⁢cos⁡u⁢du◦Solvex=3⁢sin⁡u, givenx=0andx=3, to calculate the new boundsu=0..π2◦Substitute in the values ofxanddx∫03−x2+9ⅆx=∫3⁢−9⁢sin⁡u2+9⁢cos⁡uⅆu◦Evaluate roots∫3⋅3⁢cos⁡u⋅cos⁡uⅆu◦Multiply3⋅3⁢cos⁡u⋅cos⁡u∫9⁢cos⁡u2ⅆu◦ApplyPythagorastrig identity,cos⁡u2=1−sin⁡u2∫9⁢1−sin⁡u2ⅆu◦Evaluate∫−9⁢sin⁡u2+9ⅆu◦This simplifies to∫0π2−9⁢sin⁡u2+9ⅆu▫2. Apply thesumrule◦Recall the definition of thesumrule∫f⁡u+g⁡uⅆu=∫f⁡uⅆu+∫g⁡uⅆuf⁡u=−9⁢sin⁡u2g⁡u=9This gives:∫0π2−9⁢sin⁡u2ⅆu+∫0π29ⅆu▫3. Apply theconstant multiplerule to the term∫−9⁢sin⁡u2ⅆu◦Recall the definition of theconstant multiplerule∫C⁢f⁡uⅆu=C⁢∫f⁡uⅆu◦This means:∫−9⁢sin⁡u2ⅆu=−9⁢∫sin⁡u2ⅆuWe can rewrite the integral as:−9⁢∫0π2sin⁡u2ⅆu+∫0π29ⅆu▫4. Rewrite◦Equivalent expressionsin⁡u2=12−cos⁡2⁢u2This gives:−9⁢∫0π212−cos⁡2⁢u2ⅆu+∫0π29ⅆu▫5. Apply thesumrule◦Recall the definition of thesumrule∫f⁡u+g⁡uⅆu=∫f⁡uⅆu+∫g⁡uⅆuf⁡u=12g⁡u=−cos⁡2⁢u2This gives:−9⁢∫0π212ⅆu+∫0π2−cos⁡2⁢u2ⅆu+∫0π29ⅆu▫6. Apply theconstantrule to the term∫12ⅆu◦Recall the definition of theconstantrule∫Cⅆu=C⁢u◦This means:∫12ⅆu=u2We can now rewrite the integral as:−9⁢π4−9⁢∫0π2−cos⁡2⁢u2ⅆu+∫0π29ⅆu▫7. Apply theconstant multiplerule to the term∫−cos⁡2⁢u2ⅆu◦Recall the definition of theconstant multiplerule∫C⁢f⁡uⅆu=C⁢∫f⁡uⅆu◦This means:∫−cos⁡2⁢u2ⅆu=−∫cos⁡2⁢uⅆu2We can rewrite the integral as:−9⁢π4−9⁢−∫0π2cos⁡2⁢uⅆu2+∫0π29ⅆu▫8. Apply a change of variables to rewrite the integral in terms ofv◦Letvbev=2⁢u◦Differentiate both sidesdv=2⁢du◦Isolate equation fordudu=dv2◦Solvev=2⁢u, givenu=0andu=π2, to calculate the new boundsv=0..π◦Substitute the values for v and du back into the original∫0π2cos⁡2⁢uⅆu=∫0πcos⁡v2ⅆvThis gives:−9⁢π4+9⁢∫0πcos⁡v2ⅆv2+∫0π29ⅆu▫9. Apply theconstant multiplerule to the term∫cos⁡v2ⅆv◦Recall the definition of theconstant multiplerule∫C⁢f⁡vⅆv=C⁢∫f⁡vⅆv◦This means:∫cos⁡v2ⅆv=∫cos⁡vⅆv2We can rewrite the integral as:−9⁢π4−9⁢−∫0πcos⁡vⅆv22+∫0π29ⅆu▫10. Evaluate the integral ofcos(v)◦Recall the definition of thecosrule∫cos⁡vⅆv=sin⁡v◦Apply limits of definite integralsin⁡vv=π|sin⁡vv=π−sin⁡vv=0|sin⁡vv=0This gives:−9⁢π4+∫0π29ⅆu▫11. Apply theconstantrule to the term∫9ⅆv◦Recall the definition of theconstantrule∫Cⅆv=C⁢v◦This means:∫9ⅆv=9⁢vWe can now rewrite the integral as:9⁢π4

(6)

 

Smarter Reverting of a Change of Variables

Expansion will not occur in some cases when reverting a change of variables. This can produce expected simpler results.

You can see this, for example, in the case of a factored polynomial. In the example below the result of Step 3 is no longer expanded.

ShowSolutionIntx⋅x2−15, x 

Integration Steps∫x⁢x2−15ⅆx▫1. Apply a change of variables to rewrite the integral in terms ofu◦Letubeu=x2−1◦Differentiate both sidesdu=2⁢x⁢dx◦Isolate equation fordxdx=12⁢x⋅du◦Substitute the values for u and dx back into the original∫x⁢x2−15ⅆx=∫u5ⅆu2This gives:∫u5ⅆu2▫2. Apply thepowerrule to the term∫u5ⅆu◦Recall the definition of thepowerrule, for n≠-1∫unⅆu=un+1n+1◦This means:∫u5ⅆu=u5+15+1◦So,∫u5ⅆu=u66We can rewrite the integral as:u612▫3. Revert change of variable◦Variable we defined in step1u=x2−1This gives:112⋅x2−16•Add constant of integrationx2−1612+C

(7)

Matrix Transpose

The transpose of a matrix is a straightforward linear algebra operation once you know the process, which is now easily accessible through the TransposeSteps command.

 

withStudent:-LinearAlgebra:

M ≔ 111221223132:

TransposeStepsM

M=111221223132•Recall the definition of transpose. Given:A=A1A2B1B2C1C2•The rows are switched to columns.AT=A1B1C1A2B2C2•Take the first row of the given matrix.1112........•Insert it as the first column of the result matrix.11....12....•Follow the pattern to copy all rows to columns. The transpose is:MT=112131122232

(8)

 

Improved Ability to Generate Similar Problems

• 

Instructors and students have long wished for an easy way to generate questions similar to the ones they’re working on. Depending on the question, this isn’t always straightforward. In previous versions of Maple, we took our first step toward solving this with the GenerateSimilar command, which generates a new random expression with properties similar to a given one. For example, if you input a quadratic with integer coefficients and integer roots, GenerateSimilar will produce a new expression that shares those characteristics.

• 

With Maple 2025, we’ve expanded this capability to include derivatives, making it even easier for students to practice differentiation! Given a %diff expression, it will produce a new %diff that requires the same core steps required to find the solution.

> 

RandomTools:-GenerateSimilar(  %diff( 6/sqrt(z^3) + 1/(8*z^4) - 1/(3*z^10), z ) );

ⅆⅆz9−6⁢z3−6⁢z2+25⁢z4−14⁢z10

(9)
• 

Here is an example that requires the use of the Chain Rule to solve:

> 

expr := %diff( sin(5*x), x );

expr≔ⅆⅆxsin⁡5⁢x

(10)
> 

Student:-Calculus1:-ShowSolution( expr );

Differentiation Stepsⅆⅆxsin⁡5⁢x▫1. Apply thechainrule to the termsin⁡5⁢x◦Recall the definition of thechainruleⅆⅆxf⁡g⁡x=f'⁡g⁡x⁢ⅆⅆxg⁡x◦Outside functionf⁡v=sin⁡v◦Inside functiong⁡x=5⁢x◦Derivative of outside functionⅆⅆvf⁡v=cos⁡v◦Apply compositionf'⁡g⁡x=cos⁡5⁢x◦Derivative of inside functionⅆⅆxg⁡x=5◦Put it all togetherⅆⅆxf⁡g⁡x⁢ⅆⅆxg⁡x=cos⁡5⁢x⋅5This gives:5⁢cos⁡5⁢x

(11)
• 

Generate a new expression that will also require the Chain Rule at some step in the solution:

> 

newexpr := RandomTools:-GenerateSimilar(expr);

newexpr≔ⅆⅆx6⁢cos⁡−4⁢x−9

(12)
> 

Student:-Calculus1:-ShowSolution( newexpr );

Differentiation Stepsⅆⅆx6⁢cos⁡4⁢x+9▫1. Apply theconstant multiplerule to the termⅆⅆx6⁢cos⁡4⁢x+9◦Recall the definition of theconstant multipleruleⅆⅆxC⁢f⁡x=C⁢ⅆⅆxf⁡x◦This means:ⅆⅆx6⁢cos⁡4⁢x+9=6⋅ⅆⅆxcos⁡4⁢x+9We can rewrite the derivative as:6⁢ⅆⅆxcos⁡4⁢x+9▫2. Apply thechainrule to the termcos⁡4⁢x+9◦Recall the definition of thechainruleⅆⅆxf⁡g⁡x=f'⁡g⁡x⁢ⅆⅆxg⁡x◦Outside functionf⁡v=cos⁡v◦Inside functiong⁡x=4⁢x+9◦Derivative of outside functionⅆⅆvf⁡v=−sin⁡v◦Apply compositionf'⁡g⁡x=−sin⁡4⁢x+9◦Derivative of inside functionⅆⅆxg⁡x=4◦Put it all togetherⅆⅆxf⁡g⁡x⁢ⅆⅆxg⁡x=−sin⁡4⁢x+9⋅4This gives:−24⁢sin⁡4⁢x+9

(13)
• 

For convenience an option has been added to facilitate the creation of a list of random expressions instead of just one.  The following example creates 5 new expressions at once.

> 

RandomTools:-GenerateSimilar(  %diff( (y-4)*(2*y+y^2), y), repeat=5);

ⅆⅆy−3⁢y+8⁢y⁢y+10,ⅆⅆy−5⁢y−1⁢y⁢y+9,ⅆⅆy−8⁢y−7⁢y⁢−y+3,ⅆⅆy−−7⁢y−4⁢y⁢y+9,ⅆⅆy−−4⁢y+5⁢y⁢y+3

(14)

Try Another: A New Addition to Check My Work

Maplesoft’s Check My Work feature has been widely popular among students and educators, with many saying, "I wish I had this when I was a student!" It helps learners pinpoint exactly where they went wrong -- an invaluable tool for building understanding.

With Maple 2025, we’re taking it a step further. Imagine a student working through a problem, making a mistake, correcting it, but still needing another question to reinforce their understanding. That’s where the new Try Another button comes in—allowing students to generate a similar question instantly and keep practicing until they feel confident.

This enhancement is now available across all practice commands—Factor Practice, Simplify Practice, Integration Practice and more—leveraging GenerateSimilar to provide an endless stream of practice opportunities. Educators can also integrate this feature with Maple Learn scripting tools, making it even easier to create interactive practice sheets.

With Maple 2025, students don’t just find their mistakes—they get the extra practice they need to truly master the material.

 

  Grading:-FactorPractice2⋅x3−8⋅x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Grading:-SimplifyPractice⁡cos⁡x3+cos⁡x⁢sin⁡x2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Grading:-LimitPractice Limit sinxx,x=0 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The new Try-Another button is also available when the activity is deployed to Maple Learn.  Students can practice anywhere in their web-browser using a link generated by any of these "Practice" commands:

Grading:-FactorPractice2⋅x3−8⋅x,output=link 

https://learn.maplesoft.com/d/IQCRFUBULGAKHKBQHKJNANJNJIKNASGNHILKBNIFMPJLGNBKFSLNEMEOOGAMHUJPBLFHFHCFPFPNCMKOEPMPDIETENBOOFNULONK

Feedback Output Options

Maple 2025 adds new options output=matrix and output=table to return or show tabular pairs of feedback.  This is useful for viewing Check-My-Work feedback on answers imported from another system, or manually entered.   The student's steps are provided as a list of expressions, then feedback is generated and displayed line-by-line next to each expression.   The feedback results can be visually processed, or used as a data-generation step in a larger program allowing you to add customized analysis.

withGrading: 

feedback  ≔  SolveFeedback 2 x−1=4, 2 x=5,x=52 , 'output'=matrix:

feedback3

x=52Good job, this is the correct solution

(15)

 SolveFeedback 2 x−1=4, 2 x=5,x=52 , 'output'=table:

2⁢x−1=4

 

2⁢x=5

Ok

x=52

Good job, this is the correct solution

``

Great! You found the correct solution

 

Student:-ODEs

The Student:-ODEs subpackage includes many improvements for 2025.

• 

Constants of integration are now in the more friendly form c__i instead of _C1, _C2, ... as they were in previous versions.  

• 

ODE solutions are now better simplified and the constants of integrations are redefined as necessary to achieve the most compact form.  

• 

Polynomial particular solutions for 2nd order linear ODEs which are found via the series method are now used to completely solve the ODE via reduction of order.

• 

For some ODEs, solutions involving non-real constants had been ignored; those are now included.

• 

In some examples, answers no longer include redundant solutions.

• 

Additional improvements have also been made.

• 

The ODEPlot command can now be called with an input system.

Improved solutions

Here are a few examples of some improved solutions:

  withStudent:-ODEs: 
In this example the factor yx is now ignored because it does not include any derivatives, leading to a simpler solution:

ODEStepsyx⋅diffyx,x=0

Let's solvey⁡x⁢ⅆⅆxy⁡x=0•Highest derivative means the order of the ODE is1ⅆⅆxy⁡x•Separate variablesⅆⅆxy⁡x=0•Integrate both sides with respect tox∫ⅆⅆxy⁡xⅆx=∫0ⅆx+c__1•Evaluate integraly⁡x=c__1

(16)

In this example the preliminary solution now undergoes extra simplification and the constants of integration are redefined, leading to a simpler result:
ODEStepsdiffyx,x−ln37⋅ yx = −3 

Let's solveⅆⅆxy⁡x−ln⁡3⁢y⁡x7=−3•Highest derivative means the order of the ODE is1ⅆⅆxy⁡x•Solve for the highest derivativeⅆⅆxy⁡x=ln⁡3⁢y⁡x7−3•Separate variablesⅆⅆxy⁡xln⁡3⁢y⁡x7−3=1•Integrate both sides with respect tox∫ⅆⅆxy⁡xln⁡3⁢y⁡x7−3ⅆx=∫1ⅆx+c__1•Evaluate integral7⁢ln⁡ln⁡3⁢y⁡x−21ln⁡3=x+c__1•Solve fory⁡xy⁡x=ⅇc__1⁢ln⁡37+x⁢ln⁡37+21ln⁡3•Simplifyy⁡x=3c__17+x7+21ln⁡3•Redefine the integration constant(s)y⁡x=c__1⁢3x7+21ln⁡3

(17)

For separable equations, after integrating both sides, Maple previously could get stuck trying to make the resulting solution explicit. There have been some adjustments to help avoid this problem, which have led to this new result as an example:
ODESteps(5⋅t2+1⋅diffyt, t = 4⋅t⋅yt⋅(y(t)3−1))

Let's solve5⁢t2+1⁢ⅆⅆty⁡t=4⁢t⁢y⁡t⁢y⁡t3−1•Highest derivative means the order of the ODE is1ⅆⅆty⁡t•Solve for the highest derivativeⅆⅆty⁡t=4⁢t⁢y⁡t⁢y⁡t3−15⁢t2+1•Separate variablesⅆⅆty⁡ty⁡t⁢y⁡t3−1=4⁢t5⁢t2+1•Integrate both sides with respect tot∫ⅆⅆty⁡ty⁡t⁢y⁡t3−1ⅆt=∫4⁢t5⁢t2+1ⅆt+c__1•Evaluate integralln⁡y⁡t2+y⁡t+13+ln⁡y⁡t−13−ln⁡y⁡t=2⁢ln⁡t2+15+c__1

(18)

In this reduction of order example, the solution of the reduced ODE now undergoes extra simplification and its integration constant is redefined, leading to a simpler solution for the original ODE:
ODEStepsdiffyx,x,x + diffyx,x2⋅ x = 0

Let's solveⅆ2ⅆx2y⁡x+ⅆⅆxy⁡x2⁢x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Make substitutionu=ⅆⅆxy⁡xto reduce order of ODEⅆⅆxu⁡x+u⁡x2⁢x=0•Solve for the highest derivativeⅆⅆxu⁡x=−u⁡x2⁢x•Separate variablesⅆⅆxu⁡xu⁡x2=−x•Integrate both sides with respect tox∫ⅆⅆxu⁡xu⁡x2ⅆx=∫−xⅆx+c__1•Evaluate integral−1u⁡x=−x22+c__1•Solve foru⁡xu⁡x=−2−x2+2⁢c__1•Simplifyu⁡x=2x2−2⁢c__1•Redefine the integration constant(s)u⁡x=2x2+c__1•Solve 1st ODE foru⁡xu⁡x=2x2+c__1•Make substitutionu=ⅆⅆxy⁡xⅆⅆxy⁡x=2x2+c__1•Integrate both sides to solve fory⁡x∫ⅆⅆxy⁡xⅆx=∫2x2+c__1ⅆx+c__2•Compute integralsy⁡x=2⁢arctan⁡xc__1c__1+c__2

(19)

The result for this ODE problem now includes 4 new solutions which were previously ignored because they included complex non-real constants:
ODEStepsyx3⋅ diffyx,x3 = x3

Let's solvey⁡x3⁢ⅆⅆxy⁡x3=x3•Highest derivative means the order of the ODE is1ⅆⅆxy⁡x•Solve for the highest derivativeⅆⅆxy⁡x=xy⁡x,ⅆⅆxy⁡x=−12−I⁢32⁢xy⁡x,ⅆⅆxy⁡x=−12+I⁢32⁢xy⁡x▫Solve the equationⅆⅆxy⁡x=xy⁡x◦Separate variablesy⁡x⁢ⅆⅆxy⁡x=x◦Integrate both sides with respect tox∫y⁡x⁢ⅆⅆxy⁡xⅆx=∫xⅆx+c__1◦Evaluate integraly⁡x22=x22+c__1◦Solve fory⁡xy⁡x=x2+2⁢c__1,y⁡x=−x2+2⁢c__1◦Redefine the integration constant(s)y⁡x=x2+c__1,y⁡x=−x2+c__1▫Solve the equationⅆⅆxy⁡x=−12−I⁢32⁢xy⁡x◦Separate variablesy⁡x⁢ⅆⅆxy⁡x=−12−I⁢32⁢x◦Integrate both sides with respect tox∫y⁡x⁢ⅆⅆxy⁡xⅆx=∫−12−I⁢32⁢xⅆx+c__1◦Evaluate integraly⁡x22=−12−I⁢32⁢x22+c__1◦Solve fory⁡xy⁡x=−−2⁢I⁢x2⁢3−2⁢x2+8⁢c__12,y⁡x=−2⁢I⁢x2⁢3−2⁢x2+8⁢c__12◦Redefine the integration constant(s)y⁡x=−−2⁢I⁢x2⁢3−2⁢x2+c__12,y⁡x=−2⁢I⁢x2⁢3−2⁢x2+c__12▫Solve the equationⅆⅆxy⁡x=−12+I⁢32⁢xy⁡x◦Separate variablesy⁡x⁢ⅆⅆxy⁡x=−12+I⁢32⁢x◦Integrate both sides with respect tox∫y⁡x⁢ⅆⅆxy⁡xⅆx=∫−12+I⁢32⁢xⅆx+c__1◦Evaluate integraly⁡x22=−12+I⁢32⁢x22+c__1◦Solve fory⁡xy⁡x=−2⁢I⁢x2⁢3−2⁢x2+8⁢c__12,y⁡x=2⁢I⁢x2⁢3−2⁢x2+8⁢c__12◦Redefine the integration constant(s)y⁡x=−2⁢I⁢x2⁢3−2⁢x2+c__12,y⁡x=2⁢I⁢x2⁢3−2⁢x2+c__12•Set of solutionsy⁡x=x2+c__1,y⁡x=−−2⁢I⁢x2⁢3−2⁢x2+c__12,y⁡x=−2⁢I⁢x2⁢3−2⁢x2+c__12,y⁡x=−2⁢I⁢x2⁢3−2⁢x2+c__12,y⁡x=2⁢I⁢x2⁢3−2⁢x2+c__12,y⁡x=−x2+c__1

(20)

We improved the main solving algorithm, leading to a new solution for this ODE, which previously returned an error:
ODESteps3⋅ diffyx,x$2+x⋅diffyx,x−4⋅ yx=0

Let's solve3⁢ⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡x−4⁢y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=−x⁢ⅆⅆxy⁡x3+4⁢y⁡x3•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡x3−4⁢y⁡x3=0•Multiply by denominators3⁢ⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡x−4⁢y⁡x=0•Assume series solution fory⁡xy⁡x=∑k=0∞⁡ak⁢xk▫Rewrite DE with series expansions◦Convertx⋅ⅆⅆxy⁡xto series expansionx⋅ⅆⅆxy⁡x=∑k=0∞⁡ak⁢k⁢xk◦Convertⅆ2ⅆx2y⁡xto series expansionⅆ2ⅆx2y⁡x=∑k=2∞⁡ak⁢k⁢k−1⁢xk−2◦Shift index usingk->k+2ⅆ2ⅆx2y⁡x=∑k=0∞⁡ak+2⁢k+2⁢k+1⁢xkRewrite DE with series expansions∑k=0∞⁡3⁢ak+2⁢k+2⁢k+1+ak⁢k−4⁢xk=0•Each term in the series must be 0, giving the recursion relation3⁢k2+9⁢k+6⁢ak+2+ak⁢k−4=0•Recursion relation; series terminates atk=4ak+2=−ak⁢k−43⁢k2+3⁢k+2•Apply recursion relation fork=0a2=2⁢a03•Apply recursion relation fork=2a4=a218•Express in terms ofa0a4=a027•Terminating series solution of the ODE. Use reduction of order to find the second linearly independent solutiony⁡x=a0⁢1+23⁢x2+127⁢x4•Use this particular solution to reduce the order of the ODE and find a complete solutiony⁡x=1+23⁢x2+127⁢x4⁢c__1⁢∫ⅇ−x26x4+18⁢x2+272ⅆx+c__2

(21)

We improved in the first-order IVP algorithm, which previously returned an error. It now provides a new solution for this ODE.
ODEStepsdiffyx,x3 = x−22, y2=1

Let's solveⅆⅆxy⁡x3=x−22,y⁡2=1•Highest derivative means the order of the ODE is1ⅆⅆxy⁡x•Solve for the highest derivativeⅆⅆxy⁡x=x−223,ⅆⅆxy⁡x=−x−2232−I⁢3⁢x−2232,ⅆⅆxy⁡x=−x−2232+I⁢3⁢x−2232▫Solve the equationⅆⅆxy⁡x=x−223◦Integrate both sides with respect tox∫ⅆⅆxy⁡xⅆx=∫x−223ⅆx+c__1◦Evaluate integraly⁡x=3⁢x−2535+c__1◦Use initial conditiony⁡2=11=c__1◦Solve forc__1c__1=1◦Substitutec__1=1into general solution and simplifyy⁡x=1+3⁢x−6⁢x−2235◦Solution to the IVPy⁡x=1+3⁢x−6⁢x−2235▫Solve the equationⅆⅆxy⁡x=−x−2232−I⁢3⁢x−2232◦Integrate both sides with respect tox∫ⅆⅆxy⁡xⅆx=∫−x−2232−I⁢3⁢x−2232ⅆx+c__1◦Evaluate integraly⁡x=−3⁢x−253⁢1+I⁢310+c__1◦Use initial conditiony⁡2=11=c__1◦Solve forc__1c__1=1◦Substitutec__1=1into general solution and simplifyy⁡x=−3⁢x−253⁢1+I⁢310+1◦Solution to the IVPy⁡x=−3⁢x−253⁢1+I⁢310+1▫Solve the equationⅆⅆxy⁡x=−x−2232+I⁢3⁢x−2232◦Integrate both sides with respect tox∫ⅆⅆxy⁡xⅆx=∫−x−2232+I⁢3⁢x−2232ⅆx+c__1◦Evaluate integraly⁡x=3⁢x−253⁢I⁢3−110+c__1◦Use initial conditiony⁡2=11=c__1◦Solve forc__1c__1=1◦Substitutec__1=1into general solution and simplifyy⁡x=3⁢x−253⁢I⁢3−110+1◦Solution to the IVPy⁡x=3⁢x−253⁢I⁢3−110+1•Set of solutionsy⁡x=1+3⁢x−6⁢x−2235,y⁡x=−3⁢x−253⁢1+I⁢310+1,y⁡x=3⁢x−253⁢I⁢3−110+1

(22)

The first order IVP algorithm now tries to make the general solution explicit before attempting to solve for the integration constants. Instead of an error, we now get the following solution:
ODEStepsdiffyx, x − 2⋅ yx = 2⋅ sqrtyx,y0 = 1

Let's solveⅆⅆxy⁡x−2⁢y⁡x=2⁢y⁡x,y⁡0=1•Highest derivative means the order of the ODE is1ⅆⅆxy⁡x•Solve for the highest derivativeⅆⅆxy⁡x=2⁢y⁡x+2⁢y⁡x•Separate variablesⅆⅆxy⁡x2⁢y⁡x+2⁢y⁡x=1•Integrate both sides with respect tox∫ⅆⅆxy⁡x2⁢y⁡x+2⁢y⁡xⅆx=∫1ⅆx+c__1•Evaluate integralln⁡y⁡x−12+arctanh⁡y⁡x=x+c__1•Convertarctanhto 'ln'ln⁡y⁡x−12+ln⁡y⁡x+12−ln⁡1−y⁡x2=x+c__1•Solve fory⁡xy⁡x=−2⁢−ⅇ2⁢x+2⁢c__1−ⅇ2⁢x+2⁢c__1+1,y⁡x=2⁢−ⅇ2⁢x+2⁢c__1−ⅇ2⁢x+2⁢c__1+1•Redefine the integration constant(s)y⁡x=−2⁢c__1⁢ⅇ2⁢x+c__1⁢ⅇ2⁢x+1,y⁡x=2⁢c__1⁢ⅇ2⁢x+c__1⁢ⅇ2⁢x+1•Use initial conditiony⁡0=11=−2⁢c__1+c__1+1•Solve forc__1c__1=0,c__1=4•Remove solutions that don't satisfy the ODEc__1=4•Substitutec__1=4into general solution and simplifyy⁡x=−4⁢ⅇ2⁢x+4⁢ⅇ2⁢x+1•Use initial conditiony⁡0=11=2⁢c__1+c__1+1•Solve forc__1c__1=0•Remove solutions that don't satisfy the ODE∅•Solution does not satisfy initial condition•Solution to the IVPy⁡x=−4⁢ⅇ2⁢x+4⁢ⅇ2⁢x+1

(23)

In this example, the IVP algorithm now uses a more correct approach to solving for the integration constant, leading to a new solution:
ODEStepsdiffyx,x+x⋅yx=1,y0=0

Let's solveⅆⅆxy⁡x+x⁢y⁡x=1,y⁡0=0•Highest derivative means the order of the ODE is1ⅆⅆxy⁡x•Solve for the highest derivativeⅆⅆxy⁡x=−x⁢y⁡x+1•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODEⅆⅆxy⁡x+x⁢y⁡x=1•The ODE is linear; multiply by an integrating factorμ⁡xμ⁡x⁢ⅆⅆxy⁡x+x⁢y⁡x=μ⁡x•Assume the lhs of the ODE is the total derivativeⅆⅆxy⁡x⁢μ⁡xμ⁡x⁢ⅆⅆxy⁡x+x⁢y⁡x=ⅆⅆxy⁡x⁢μ⁡x+y⁡x⁢ⅆⅆxμ⁡x•Isolateⅆⅆxμ⁡xⅆⅆxμ⁡x=μ⁡x⁢x•Solve to find the integrating factorμ⁡x=ⅇx22•Integrate both sides with respect tox∫ⅆⅆxy⁡x⁢μ⁡xⅆx=∫μ⁡xⅆx+c__1•Evaluate the integral on the lhsy⁡x⁢μ⁡x=∫μ⁡xⅆx+c__1•Solve fory⁡xy⁡x=∫μ⁡xⅆx+c__1μ⁡x•Substituteμ⁡x=ⅇx22y⁡x=∫ⅇx22ⅆx+c__1ⅇx22•Evaluate the integrals on the rhsy⁡x=−I⁢π⁢2⁢erf⁡I2⁢2⁢x2+c__1ⅇx22•Simplifyy⁡x=−I⁢π⁢2⁢erf⁡I2⁢2⁢x−2⁢c__1⁢ⅇ−x222•Use initial conditiony⁡0=00=c__1•Solve forc__1c__1=0•Substitutec__1=0into general solution and simplifyy⁡x=−I2⁢π⁢2⁢erf⁡I2⁢2⁢x⁢ⅇ−x22•Solution to the IVPy⁡x=−I2⁢π⁢2⁢erf⁡I2⁢2⁢x⁢ⅇ−x22

(24)

This example is now solved after fixing a problem where the algorithm for constant coeff linear ODEs of order higher than 2 was not being used unless requested explicitly by the user:
aliasr = 108+12⋅931213,s = 108+12⋅931213:ODEStepsdiffyx,x$3+diffyx,x+yx=1

Let's solveⅆ3ⅆx3y⁡x+ⅆⅆxy⁡x+y⁡x=1•Highest derivative means the order of the ODE is3ⅆ3ⅆx3y⁡x•Characteristic polynomial of homogeneous ODEr3+r+1=0•Roots of the characteristic polynomialr=−108+12⁢93136+2108+12⁢9313,108+12⁢931312−1108+12⁢9313+I⁢3⁢−108+12⁢93136−2108+12⁢93132,108+12⁢931312−1108+12⁢9313−I⁢3⁢−108+12⁢93136−2108+12⁢93132•Homogeneous solution fromr=−108+12⁢93136+2108+12⁢9313y1⁡x=ⅇ−108+12⁢93136+2108+12⁢9313⁢x•Homogeneous solutions fromr=108+12⁢931312−1108+12⁢9313+I⁢3⁢−108+12⁢93136−2108+12⁢93132andr=108+12⁢931312−1108+12⁢9313−I⁢3⁢−108+12⁢93136−2108+12⁢93132y2⁡x=−ⅇ108+12⁢931312−1108+12⁢9313⁢x⁢sin⁡3⁢−108+12⁢93136−2108+12⁢9313⁢x2,y3⁡x=ⅇ108+12⁢931312−1108+12⁢9313⁢x⁢cos⁡3⁢−108+12⁢93136−2108+12⁢9313⁢x2•General solution of the ODEy⁡x=c__1⁢y1⁡x+c__2⁢y2⁡x+c__3⁢y3⁡x+yp⁡x•Substitute in solutions of the homogeneous ODEy⁡x=c__1⁢ⅇ−108+12⁢93136+2108+12⁢9313⁢x−c__2⁢ⅇ108+12⁢931312−1108+12⁢9313⁢x⁢sin⁡3⁢−108+12⁢93136−2108+12⁢9313⁢x2+c__3⁢ⅇ108+12⁢931312−1108+12⁢9313⁢x⁢cos⁡3⁢−108+12⁢93136−2108+12⁢9313⁢x2+yp⁡x▫Find a particular solutionyp⁡xof the ODE◦Define the forcing function of the ODEf⁡x=1◦Form of the particular solution to the ODE where theui⁡xare to be foundyp⁡x=∑i=13⁡ui⁡x⁢yi⁡x◦Calculate the 1st derivative ofyp⁡xⅆⅆxyp⁡x=∑i=13⁡ⅆⅆxui⁡x⁢yi⁡x+ui⁡x⁢ⅆⅆxyi⁡x◦Choose equation to add to a system of equations inⅆⅆxui⁡x∑i=13⁡ⅆⅆxui⁡x⁢yi⁡x=0◦Calculate the 2nd derivative ofyp⁡xⅆ2ⅆx2yp⁡x=∑i=13⁡ⅆⅆxui⁡x⁢ⅆⅆxyi⁡x+ui⁡x⁢ⅆ2ⅆx2yi⁡x◦Choose equation to add to a system of equations inⅆⅆxui⁡x∑i=13⁡ⅆⅆxui⁡x⁢ⅆⅆxyi⁡x=0◦The ODE is of the following form where thePi⁡xin this situation are the coefficients of the derivatives in the ODEⅆ3ⅆx3y⁡x+∑i=02⁡Pi⁡x⁢ⅆiⅆxiy⁡x=f⁡x◦Substituteyp⁡x=∑i=13⁡ui⁡x⁢yi⁡xinto the ODE∑j=02⁡Pj⁡x⁢∑i=13⁡ui⁡x⁢ⅆjⅆxjyi⁡x+∑i=13⁡ⅆⅆxui⁡x⁢ⅆ2ⅆx2yi⁡x+ui⁡x⁢ⅆ3ⅆx3yi⁡x=f⁡x◦Rearrange the ODE∑i=13⁡ui⁡x⋅∑j=02⁡Pj⁡x⁢ⅆjⅆxjyi⁡x+ⅆ3ⅆx3yi⁡x+ⅆⅆxui⁡x⁢ⅆ2ⅆx2yi⁡x=f⁡x◦Notice thatyi⁡xare solutions to the homogeneous equation so the first term in the sum is 0∑i=13⁡ⅆⅆxui⁡x⁢ⅆ2ⅆx2yi⁡x=f⁡x◦We have now made a system of3equations in3unknowns (ⅆⅆxui⁡x)∑i=13⁡ⅆⅆxui⁡x⁢yi⁡x=0,∑i=13⁡ⅆⅆxui⁡x⁢ⅆⅆxyi⁡x=0,∑i=13⁡ⅆⅆxui⁡x⁢ⅆ2ⅆx2yi⁡x=f⁡x◦Convert the system to linear algebra format, notice that the matrix is the wronskianWy1⁡xy2⁡xy3⁡xⅆⅆxy1⁡xⅆⅆxy2⁡xⅆⅆxy3⁡xⅆ2ⅆx2y1⁡xⅆ2ⅆx2y2⁡xⅆ2ⅆx2y3⁡x·ⅆⅆxu1⁡xⅆⅆxu2⁡xⅆⅆxu3⁡x=00f⁡x◦Solve for the varied parametersu1⁡xu2⁡xu3⁡x=∫W−1·00f⁡xⅆx◦Substitute in the homogeneous solutions and forcing function and solveu1⁡xu2⁡xu3⁡x=108+12⁢9323+r⁢93−9⁢r+12⁢9+93⁢108+12⁢9323+12216⁢3⁢93+31⁢ⅇ−108+12⁢9323−12⁢x6⁢108+12⁢9313⁢108+12⁢9313−3⁢9+93⁢36⁢−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132−36⁢−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+12⁢3⁢−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132−3⁢108+12⁢9323⁢−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132−−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+3⁢108+12⁢9323⁢−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢9313218⁢108+12⁢9313⁢3⁢93+31−3⁢9+93⁢−36⁢−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132−36⁢−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+12⁢3⁢−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132−−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+3⁢108+12⁢9323⁢−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132+3⁢108+12⁢9323⁢−r12+1108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢cos⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢93132−−3⁢r12−3108+12⁢9313⁢ⅇx⁢−r12+1108+12⁢9313⁢sin⁡x⁢3⁢r12+3108+12⁢9313−r12+1108+12⁢93132+3⁢r12+3108+12⁢9313218⁢108+12⁢9313⁢3⁢93+31Find a particular solutionyp⁡xof the ODEyp⁡x=1•Substitute particular solution into general solution to ODEy⁡x=c__1⁢ⅇ−108+12⁢93136+2108+12⁢9313⁢x−c__2⁢ⅇ108+12⁢931312−1108+12⁢9313⁢x⁢sin⁡3⁢−108+12⁢93136−2108+12⁢9313⁢x2+c__3⁢ⅇ108+12⁢931312−1108+12⁢9313⁢x⁢cos⁡3⁢−108+12⁢93136−2108+12⁢9313⁢x2+1

(25)

New ODEPlot calling sequence

Maple 2025 introduces a major enhancement to the ODEPlot command, which generates direction fields and solution trajectories for systems of two first-order ODEs. Previously, ODEPlot did not support input arguments, meaning you could only generate a default plot of an ODE system. To modify the system or adjust plotting parameters, you had to manually tweak the interactive controls.

In Maple 2025, we've extended the ODEPlot command to allow direct input of a custom system via the command line, making it much more convenient to quickly visualize a given system. The new calling sequence is as follows:

ODEPlot(system, dependent_variable_ranges, independent_variable_range, parameter_values, initial_values);

For this calling sequence only the first argument system is required, and the rest are optional.

Let's see how this works with a simple example:
withStudent:-ODEs:
sys ≔ diff⁡x⁡t,t=exp⁡−y⁡t,diff⁡y⁡t,t=k−x⁡t;

sys≔ⅆⅆtx⁡t=ⅇ−y⁡t,ⅆⅆty⁡t=k−x⁡t

(26)
> 

ODEPlotsys; 

Select a system of differential equations

 

 

 

Ranges

 

 

from

to

from

to

from

to

 

Parameter Values

 

 

 

 

 

 

 

 

 

 

 

The system has been plotted and a number of default choices have been made for the parameter value and the variable ranges and initial points. If instead we'd like to specify those as well, we can do it as follows:

> 

t_range ≔ t=0..5;

t_range≔t=0..5

(27)
> 

x_range ≔ x = −3..3;

x_range≔x=−3..3

(28)
> 

y_range ≔ y = −2..2; 

y_range≔y=−2..2

(29)
> 

parameter_values ≔ k=−1;

parameter_values≔k=−1

(30)
> 

initial_values ≔ t,x,y=0,1,2,0,−3,−2,0,−2,0; 

initial_values≔t,x,y=0,1,2,0,−3,−2,0,−2,0

(31)
> 

ODEPlotsys,  t_range, x_range, y_range, parameter_values, initial_values;  

Select a system of differential equations

 

 

 

Ranges

 

 

from

to

from

to

from

to

 

Parameter Values

 

 

 

 

 

 

 

 

 

 

 

Radius of Convergence

Maple 2025 includes two new commands for checking if an infinite series converges unconditionally and for finding its radius of convergence, Converges and ConvergenceRadius.  These are part of the SumTools:-DefiniteSum package.withSumTools:-DefiniteSum

ConvergenceRadius,Converges,CreativeTelescoping,Definite,SummableSpace,Telescoping,pFqToStandardFunctions

(32)

Converges∑n=1∞1n

false

(33)

 Converges∑n=1∞1n2

true

(34)

The exponential series converges unconditionally.

Converges∑n=1∞xnn!

true

(35)

Converges∑n=1∞xnn 

FAIL

(36)

 Converges∑n&equals;1∞xnn assuming −1<x<1

true

(37)

 ConvergenceRadius∑n&equals;1∞xnn&comma;x

x<1

(38)

 ConvergenceRadius∑n&equals;1∞2 nnxn&comma;x

x<14

(39)

 

Find a Formula for the nth Term of a Given Integer Sequence

• 

Have you ever struggled to find the next number in a sequence or wondered what formula defines it? With the new IdentifySequence command in Maple 2025, you don’t have to guess! Simply input a sequence of integers, and IdentifySequence will attempt to identify a formula for the nth term—unlocking patterns and insights instantly.

> 

IdentifySequence([1, 3, 5, 7, 9], 'n');

2⁢n−1

(40)
> 

IdentifySequence([1, 1, 2, 3, 5, 8, 13],'x');

12−510⁢−52+12x−1+12+510⁢52+12x−1

(41)
> 

IdentifySequence([1, 2, 4, 8, 16], 't', 'start' = 0);

2t

(42)
> 

IdentifySequence([88, 440, 2200, 11000],'t');

88⁢5t−1

(43)
> 

IdentifySequence([19,15,11,7,3], 'n');

−4⁢n+23

(44)