Calculators Solving ODEs on the TI-89: Tips & Tricks

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The TI-89 can solve certain types of ordinary differential equations (ODEs), but it is limited to second-degree equations. Users have found that reformatting equations, such as converting higher-order ODEs into a system of first-order equations, can help in finding solutions. For example, expressing y''' + 3y'' + 2y' - 5y = sin(2t) in terms of y'1, y'2, and y'3 can facilitate input into the calculator. The calculator's preference for expressing tangent and cotangent in terms of sine and cosine can complicate simpler equations. Overall, while the TI-89 has its limitations, users are exploring various methods to maximize its capabilities for solving ODEs.
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1. Is it possible for the TI-89 to solve Exact Equations?

Ex: (2x-1)dx + (3y+7)dy = 0

I've tried various forms of input, but I cannot find a way for the Calculator to give me a complete answer. My best luck so far was:

(2x-1)x' + (3y+7)y' = 0. The y' part was correct, the x' part was incomplete.

2. I've been reading about how to convert higher order equations into a form that the calculator can solve.

Ex: y''' + 3y'' + 2y' - 5y = sin(2t)

Can be written as:

y'(subscript 3) = 5y(subscript 1) - 2y(subscript 2) -3y(subscript 3) + sin(2t)

How might I enter this into the TI-89?

3. Why does the TI-89 put tan and cot in terms of sin and cos? I have seen rather simple equations turn into complicated messes because of this.

Thanks in advance.:smile:
 
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Did you try (2x-1) + (3y+7)y' = 0?

##y'_3##? Looks like an unusual notation. What is wrong with y'''?
 
Your suggestion worked, I find it odd that leaving out x' makes it work though.

The TI-89 can only do DE's up to the second degree. Would it help if I explained the method in full?

Thanks for your help.
 
Last edited:
The TI-89 can only do DE's up to the second degree.
Hmm... can it solve coupled DEs?

z=y', z''+3z'+2z-5y=sin(2t)

Your suggestion worked, I find it odd that leaving out x' makes it work though.
dx/dx=1
 
mfb said:
Hmm... can it solve coupled DEs?

z=y', z''+3z'+2z-5y=sin(2t)

No, it can't. I see what you're doing though, and its very similar to the method I am trying to use. In my method:

y1 = y, y2 = y', y3 = y''...yn = y ^(n-1).

From these:

y'1 = y' = y2, y'2 = y'' = y3...y'(n) = y^n.

This gives the system:

y'1 = y2
y'2 = y3

My example from earlier:

Ex: y''' + 3y'' + 2y' - 5y = sin(2t)

Can be written as:

y'(subscript 3) = 5y(subscript 1) - 2y(subscript 2) -3y(subscript 3) + sin(2t)

I'm puzzled as to how to enter this in the calculator though.
 
I understand I could graph this, but how would that help me?
 
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