Solving a 2nd Order Differential Equation with Exponential Terms

[squaremeplz]

Homework Statement

[tex] \frac { d^2 \theta }{d x'^2 } = -y *exp(\theta) [/tex] eq. 1

mayb be integrated to yield

[tex] exp(\theta) = \frac {a}{cosh^2(b \frac{+}{-} \sqrt \frac{a*y}{2} * x')} [/tex]

[tex] \theta = f(y,x') [/tex]

Homework Equations

The Attempt at a Solution

the exponent is throwing me off, but i probably have to use the following properties

[tex] \frac { d^2 \theta }{d x'^2 } + \sqrt{y}*exp(/theta) = 0 [/tex]


[tex] g = [1/2 (e^k^x + e^-^k^x)]^2 = cosh^2(kx) [/tex]

the problem is the exp(theta), do I have to replace the variable with something like

g = ln(theta)

then find

[tex] \frac { d^2 g }{d x'^2 } [/tex]

to get

[tex] \frac { d^2 g }{d x'^2 } + \sqrt{y}*g = 0 [/tex]

then apply

m^2 + k^2 = 0?

thanks

 

[Mark44]

I'm confused. In the other thread you started, you have
[tex] \frac { \partial^2 \theta }{\partial x'^2 } = -y *exp(\theta) [/tex]
and in this one you have it as a 2nd-order ODE.
Since theta is a function of y and x', you really need partial derivatives, not ordinary derivatives. If theta were a function of a single variable, say x', then you could talk about d(theta)/dx' and d^2(theta)/dx'^2.

Having said that, how do you go from equation 1 to this equation? IOW, how did y in the first equation become sqrt(y) in this equation?
[tex] \frac { d^2 \theta }{d x'^2 } + \sqrt{y}*exp(\theta) = 0 [/tex]

 

[squaremeplz]

sorry, at first it was a partial derivative but then I realized that theta is only dependent on the spatial variable x', like you said. Since this is a steady state problem (conditions for the problem approaching a steady state are given by the rate of reaction, it reduces to a ODE

Anyway,

I found more info on how to solve the equation

[tex] x' = \int \frac{d \theta}{\sqrt{-2*\int -y*exp(\theta) d \theta}} [/tex]

then i get

[tex] x' = \int \frac {d\theta}{\sqrt{2*y*exp(\theta) + 2*y*c}} [/tex]and now I am trying to figure out the above integral.

 

[Mark44]

Factor 2y out of the terms in the radical and bring them out of the radical and out of the integral so that you have exp(theta) + c in the radical. I would then see if a substitution would work.

 

[squaremeplz]

ok so following ur suggestion i did

[tex] x' = \frac {1}{\sqrt{2y}} * \int \frac {d\theta}{\sqrt{\int exp(\theta)d\theta}} [/tex]

[tex] x' = \frac {1}{\sqrt{2y}} * \int \frac {d\theta}{\sqrt{ exp(\theta)+c}} [/tex]

then i use the identity for int 1/sqrt(theta^2 + a^2) that is

[tex] x' = \frac {1}{\sqrt{2y}} * ln | exp(\theta/2) + \sqrt{exp(\theta) + c} | + a [/tex]

but when i try to solve for theta, i don't know how to reduce the exp to fit the form of the solution the book gives for exp(theta)

 

[Mark44]

Not sure if this will help, but the integral formula you used can be expressed another way.
[tex]\int \frac{dx}{\sqrt{x^2 + a^2}}~=~sinh^{-1}(x/a) + C[/tex]

One thing I've been bothered by is treating y as if it were a constant in the integration with respect to theta. You have theta = f(y, x'), and in the other thread you said that y = x/x'.

 

[squaremeplz]

hey mike, thanks so much for your hlep. yes that identity definitely helps.

please try to ignore the other post

y here is known as the frank-kamenetskii parameter

[tex] y = \frac {E}{RT^2_a} \frac {Q}{h} *r^2*z*exp(-E/RT_a) [/tex]

now, the r^2 in the y comes from x' = x/r

this is the substitution to drop the units of x, where r can be the half width or radius of a cylinder.

specifically, by the chain rule

[tex] \Delta x' = r^2 \Delta [/tex]

Quoting the book, In the simplest problems, the boundary condition is \theta = 0 at the surface, and the critical condition reduces to [tex] y = const = y_c_r_i_t [/tex] since neither in the equation nor in the boundary condions are there any parameters other than y.

So i think its ok to factor out y, no?

 

[Mark44]

That's Mark...
You have [tex] \Delta x' = r^2 \Delta [/tex]
Don't you mean this?
[tex] \Delta x' = \Delta r^2[/tex]
Based on what you said about y, sounds like you can move it out of the integral.

 

[squaremeplz]

sorry, mark :)

I still don't get the same answer as in the book

I did

[tex] x' = \frac {1}{\sqrt{2y}} * sech^-^1 (\frac {exp(\frac {\theta}{2})}{\sqrt{c}}) + a [/tex]

when I solve for [tex] exp(\theta) [/tex] i get

[tex] exp(\theta) = c* sech^2(\sqrt{2y}*x' + a) [/tex]

then

[tex] exp(\theta) = \frac {2c}{cosh^2 (\sqrt {2y}*x' + a)} [/tex]

the only problem now are the constants, the book has

[tex] exp(\theta) = \frac {a}{cosh^2 ( \sqrt {\frac {ay}{2}}*x' \frac {+}{-} b)} [/tex]

im confused why they have +/- and why it is sqrt(ay)/2

 

[Mark44]

1/cosh^2(x) = sech^2(x)

 

[squaremeplz]

[tex] exp(\theta) = \frac {2c}{cosh^2 (\sqrt {2y}*x' + a)} [/tex]

the only problem now are the constants, the book has

[tex] exp(\theta) = \frac {a}{cosh^2 ( \sqrt {\frac {ay}{2}}*x' \frac {+}{-} b)} [/tex]

im confused why they have +/- and why it is sqrt(ay/2)

 

[Mark44]

Got me, so that looks like something you'll have to puzzle out...

 

[squaremeplz]

alrighty

well, thanks again for getting me this close. great help :)