Efficient Methods for Solving Non-Power Series Differential Equations

[Saladsamurai]

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?

 

[Kreizhn]

Are Pr and \eta constants with respect to \Theta? If so, why not just multiply though by \Theta' to get

\Theta '' + \frac12 P_r \eta \Theta' + \frac12 P_r \Theta = 0
which is now in a form that can be easily solved using the trial function \Theta(t) = e^{mt}.

 

[LCKurtz]

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:

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

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

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

 

[Saladsamurai]

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:

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

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

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

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

Thanks again!