Efficient Methods for Solving Non-Power Series Differential Equations

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Homework Help Overview

The discussion revolves around solving a differential equation without using power series. The equation involves terms related to variables and constants, prompting participants to explore alternative methods for finding solutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation by dividing through by Θ' and considers integrating the resulting terms. Some participants question whether certain parameters are constants with respect to Θ, suggesting alternative forms of the equation that might be easier to solve. Others share results from computational tools like Maple and Wolfram, indicating different potential solutions based on varying interpretations of the equation.

Discussion Status

The discussion is active, with participants exploring various interpretations and methods. Some have provided insights from computational tools, while others are considering the implications of constants and variable dependencies in the equation. There is no explicit consensus yet, but multiple lines of reasoning are being examined.

Contextual Notes

There is uncertainty regarding the nature of the parameters Pr and η, and whether they are constants in the context of the equation. Additionally, there are indications of possible typos in the original equation, which may affect the approaches being discussed.

Saladsamurai
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Homework Statement



I need to find a way to solve this without using power series:

Θ"/Θ' + .5Pr*η - .5Pr*Θ/Θ' = 0​




The Attempt at a Solution



I have one idea and that is to divide through by Θ' to get

Θ"/Θ' + .5Pr*η - .5Pr*Θ/Θ' = 0​

The first term Θ"/Θ' = ln[ Θ' ]'

So if I could re write the last term in a similar fashion, I could integrate directly ... but I am losing a lot of time just guessing and checking.

Any thoughts on methods?
 
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Are Pr and [itex]\eta[/itex] constants with respect to [itex]\Theta[/itex]? If so, why not just multiply though by [itex]\Theta'[/itex] to get

[tex]\Theta '' + \frac12 P_r \eta \Theta' + \frac12 P_r \Theta = 0[/tex]
which is now in a form that can be easily solved using the trial function [itex]\Theta(t) = e^{mt}[/itex].
 
Assuming this is a followup from your other thread and the second term has an independent variable in it, I plugged this equation into Maple:

y'' + axy' + ay = 0

It gave this:

[tex]y=\frac{C_1erf(\frac{x\sqrt{{-2a}}}{2})<br /> +C_2}{e^{\frac 1 2 ax^2}}[/tex]

where erf is the error function (see http://en.wikipedia.org/wiki/Error_function).

[Edit] On the other hand if you have a typo and meant

y'' + axy' - ay = 0 then Maple gives

[tex]C_1x + C_2\left(-e^{-\frac {ax^2}{2}}-x\sqrt{\frac{\pi a}{2}}erf(x\sqrt{\frac{a}{2}}) \right)[/tex]
 
Last edited:
LCKurtz said:
Assuming this is a followup from your other thread and the second term has an independent variable in it, I plugged this equation into Maple:

y'' + axy' + ay = 0

It gave this:

[tex]y=\frac{C_1erf(\frac{x\sqrt{{-2a}}}{2})<br /> +C_2}{e^{\frac 1 2 ax^2}}[/tex]

where erf is the error function (see http://en.wikipedia.org/wiki/Error_function).

[Edit] On the other hand if you have a typo and meant

y'' + axy' - ay = 0 then Maple gives

[tex]C_1x + C_2\left(-e^{-\frac {ax^2}{2}}-x\sqrt{\frac{\pi a}{2}}erf(x\sqrt{\frac{a}{2}}) \right)[/tex]

Hi LC! I plugged it into Wolfram and got the second one. This lead me to "guess" that y1 = cx and now use reduction of order to generate a second solution. I will post back with results/questions :smile:

Thanks again!
 

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