# Homework Help: 2nd order ODE

1. Nov 23, 2009

### squaremeplz

1. The problem statement, all variables and given/known data

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

mayb be integrated to yield

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

$$\theta = f(y,x')$$

2. Relevant equations

3. The attempt at a solution

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

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

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

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

g = ln(theta)

then find

$$\frac { d^2 g }{d x'^2 }$$

to get

$$\frac { d^2 g }{d x'^2 } + \sqrt{y}*g = 0$$

then apply

m^2 + k^2 = 0?

thanks

2. Nov 23, 2009

### Staff: Mentor

I'm confused. In the other thread you started, you have
$$\frac { \partial^2 \theta }{\partial x'^2 } = -y *exp(\theta)$$
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?
$$\frac { d^2 \theta }{d x'^2 } + \sqrt{y}*exp(\theta) = 0$$

3. Nov 24, 2009

### 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,

$$x' = \int \frac{d \theta}{\sqrt{-2*\int -y*exp(\theta) d \theta}}$$

then i get

$$x' = \int \frac {d\theta}{\sqrt{2*y*exp(\theta) + 2*y*c}}$$

and now im trying to figure out the above integral.

Last edited: Nov 24, 2009
4. Nov 24, 2009

### Staff: Mentor

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.

5. Nov 24, 2009

### squaremeplz

ok so following ur suggestion i did

$$x' = \frac {1}{\sqrt{2y}} * \int \frac {d\theta}{\sqrt{\int exp(\theta)d\theta}}$$

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

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

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

but when i try to solve for theta, i dont know how to reduce the exp to fit the form of the solution the book gives for exp(theta)

6. Nov 24, 2009

### Staff: Mentor

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

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'.

7. Nov 24, 2009

### 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

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

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

$$\Delta x' = r^2 \Delta$$

Quoting the book, In the simplest problems, the boundary condition is \theta = 0 at the surface, and the critical condition reduces to $$y = const = y_c_r_i_t$$ 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?

Last edited: Nov 24, 2009
8. Nov 24, 2009

### Staff: Mentor

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

9. Nov 24, 2009

### squaremeplz

sorry, mark :)

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

I did

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

when I solve for $$exp(\theta)$$ i get

$$exp(\theta) = c* sech^2(\sqrt{2y}*x' + a)$$

then

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

the only problem now are the constants, the book has

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

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

Last edited: Nov 24, 2009
10. Nov 24, 2009

### Staff: Mentor

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

11. Nov 24, 2009

### squaremeplz

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

the only problem now are the constants, the book has

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

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

12. Nov 24, 2009

### Staff: Mentor

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

13. Nov 24, 2009

### squaremeplz

alrighty

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