# A Nonlinear Second Order Differential Equation Problem

1. Nov 7, 2009

### overmage

Hello, I am a first year engineering undergraduate student, and this is my question.

1. The problem statement, all variables and given/known data
A dust particle of negligible mass starts to fall, t=0, under the influence of gravitational force through mist of saturated water vapor. The vapor condenses onto the dust particle at a constant rate of $$\lambda$$ kilogram per meter of traveled distance. The dust particle thereby develops into a rain droplet with increasing mass.

a) Calculate the acceleration of the droplet as a function of its velocity and covered distance.

b) Determine the equation of motion of the droplet by integrating the expression of the acceleration, neglecting friction, collisions, etc.

2. Relevant equations
I have already solved a). The equation (which I am told is correct) is:
$$\frac{dv}{dt} + \frac{v^{2}}{x} =$$ g

where g is the gravitational field strength.

3. The attempt at a solution
I was given the 'hint' of substituting $$f(x) = At^{n}$$ to solve this equation for the position x, and determine A and n.

I have no idea how to determine A and n, frankly. In fact, I have never done Second Order ODEs before, let alone nonlinear Second Order ODEs. (Before you ask why I would be set a question on Second Order ODEs not having done it before - my professor - let's just say I've had better elementary schools teachers.)

I arrive at this equation:

(2n$$^{2}-n)At^{n-2} =$$ g

What do I do from here now? Four hours of rearranging to no avail.

P.S. After sleuthing around, I discovered that the usual method for solving such equations is to substitute it to a first order one. This doesn't look like it...

Last edited: Nov 7, 2009
2. Nov 8, 2009

bump =(

3. Nov 8, 2009

### ehild

Hi Overmage,

$$\frac{dv}{dt} + \frac{v^{2}}{x} = g$$

is correct, it was a good job to derive it!

Now it is a first order differential equation (contains first derivative only), but non-linear. Luckily, it does not contain the time itself. There is a trick for such equations, often applied in Mechanics: We will use x as independent variable instead of t.

$$\frac{dv}{dt} =\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

Substituting for dv/dt, your equation becomes:

$$\frac{dv}{dx}v+\frac{v^{2}}{x} = g$$

Now the second trick comes:

$$\frac{dv}{dx}v=0.5\frac{d(v^2)}{dx}$$

and we choose v^2 as dependent variable, say y=v^2.

Now we arrived at a totally different-looking equation:

$$0.5 \frac{dy}{dx}+\frac{y}{x} = g$$

This is a linear equation! You can solve it by assuming y in a product form: y=F*G.

Have you learnt that method from Maths? If you do not know it yet, I'll explain.

Anyway: the solution will be:

$$v^2=\frac{2gx}{3}$$

from where you get the acceleration, and it will be a surprise!

ehild

4. Nov 8, 2009

### overmage

No, I'm not familiar with that. =( could you give me a general idea of how that works out?

5. Nov 9, 2009

### ehild

You have a differential equation of form

a(x)y'+b(x)y=c(x) (the prime means differentiation), with initial condition y=0 at x=0, the prime means differentiation.

Assume Y=F(X)G(x).

a(x)(F'G+FG')+b(x)FG =c(x). *

You have the freedom to choose G so that

a(x)FG' + b(x) FG =0

Divide the eq. by F.

a(x)G'+b(X)G=0.

You can solve this equation by separating the variables.

$$\int{\frac{dG}{G}}=\int{\frac{-b(x)}{a(x)}dx}$$

Replace back G(x) into eq. *.

a(x)F'G(x) = c(x)

$$F(x)=\int{\frac{c(x)}{a(x)G(x)}}+const$$

y=F(x)G(x).
-----------------------------------------------------------------------

0.5 y' +y/x =g ---> y'+2y/x=2g

a=1, b=2/x, c=g

$$\int{\frac{dG}{G}}=-\int{\frac{2}{x}dx} \rightarrow G(x)=x^{-2}$$

$$F'G(x)=c(x)\rightarrow F'=2gx^2 \rightarrow F(x)=\frac{2gx^3}{3}+C$$

$$y(x)= F(x)G(x)=\frac{2gx}{3}+\frac{C}{x^2}$$

C=0 according to the initial condition.

$$y(x)=\frac{2gx}{3}$$

ehild

6. Nov 9, 2009

### overmage

Hey,

Thanks alot for the help! I really appreciate it. However, can i clarify just one last thing:

shouldn't c be = 2g?

Thanks for the help you've given! Wow, this problem is kinda hard. Think you've pretty much covered it all though.

Last edited: Nov 9, 2009
7. Nov 9, 2009

### ehild

Yes, of course... :)

ehild