Solving a 2nd Order Differential Equation with Exponential Terms

• squaremeplz
In summary, the book says that the equation for exp(theta) is cosh^2(x)exp(\theta), where c is a constant. The book also says that the equation can be solved for exp(\theta) by using the substitution y = c*sech^2(\sqrt{2y}*x' + a).
squaremeplz

Homework Statement

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

mayb be integrated to yield

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

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

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

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

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 I am trying to figure out the above integral.

Last edited:
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.

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 don't know how to reduce the exp to fit the form of the solution the book gives for exp(theta)

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'.

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:
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.

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

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

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)

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

alrighty

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

1. What is a 2nd order differential equation with exponential terms?

A 2nd order differential equation with exponential terms is an equation that contains a second derivative (or second order) of a function, along with exponential terms in the form of e^x or Ae^(Bx), where A and B are constants.

2. How do you solve a 2nd order differential equation with exponential terms?

To solve a 2nd order differential equation with exponential terms, you can use the method of undetermined coefficients or the method of variation of parameters. The method of undetermined coefficients involves finding a particular solution based on the form of the exponential terms, while the method of variation of parameters involves finding a solution using a general form of the solution and then substituting it into the equation to solve for the parameters.

3. What are the initial conditions for solving a 2nd order differential equation?

The initial conditions for solving a 2nd order differential equation are the values of the function and its first derivative at a specific point. These conditions are typically provided as part of the problem and are used to determine the constants in the general solution.

4. Can a 2nd order differential equation with exponential terms have complex solutions?

Yes, a 2nd order differential equation with exponential terms can have complex solutions if the constants in the general solution are complex numbers. This is because the exponential function can have complex values for certain inputs.

5. Are there any practical applications of solving 2nd order differential equations with exponential terms?

Yes, there are many practical applications of solving 2nd order differential equations with exponential terms. These equations are commonly used in physics, engineering, and other scientific fields to model systems that involve exponential growth or decay. For example, they can be used to model population growth, radioactive decay, and the behavior of electrical circuits.

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