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Examples of differential equations

A simple Ordinary Differential Equation

The simplest differential equations are ordinary, linear differential equations of the first order with constant coefficients. For example:

<math>\frac{dy}{dt} + f(t) y = 0,</math>

where <math>f(t)</math> is some known function. We may solve this simply by rearranging it (using the chain rule) as

<math>\frac{d(e^{F(t)}y)}{dt} = 0,</math>

where <math>F(t) = \int f(t)\,dt</math>. Integrating this, we have

<math>y = A e^{-F(t)}</math>

where <math>A</math> is an arbitrary constant. (We can easily check this is a solution)

Some elaboration is needed since <math>f(t)</math> is not in fact a constant, indeed it might not even be integrable. Arguably, one must also assume something about the domains of the functions involved before the equation is fully defined. Are we talking complex functions, or just real, for example? The usual textbook approach is to discuss forming the equations well before considering how to solve them.

A simple mathematical model

Suppose a mass is attached to a spring, which exerts an attractive force on the mass proportional to the extension/compression of the spring and ignore any other forces (gravity, friction etc). We shall write the extension of the spring at a time <math>t</math> as <math>x(t)</math>. Now, using Newton's second law we can write (using convenient units)

<math>\frac{d^2x}{dt^2} = - x</math>

If we look for solutions that have the form <math>C e^{kt}</math>, where <math>C</math> is a constant, we discover the relationship <math>k^2 + 1 = 0</math>, and thus <math>k</math> must be one of the complex numbers <math>i</math> or <math>-i</math>. Thus, using Euler's theorem we can say that the solution must be of the form:

<math>x(t) = A \cos t + B \sin t</math>

To fix the unknown constants <math>A</math> and <math>B</math>, we need initial conditions, i.e. to specify the state of the system at a given time (usually taken to be <math>t = 0</math>).

For example, if we suppose at <math>t = 0</math> the extension is a unit distance (<math>x = 1</math>), and the particle is not moving (<math>dx/dt = 0</math>). We have

<math>x(0) = A \cos 0 + B \sin 0 = A = 1,</math>

and so <math>A = 1</math>.

<math>x'(0) = -A \sin 0 + B \cos 0 = B = 0,</math>

and so <math>B = 0</math>.

Therefore <math>x(t) = \cos t</math>. (This is an example of simple harmonic motion)

Improving our model

The above model of an oscillating mass on a spring is plausible but not really realistic. For a start, we've invented a perpetual motion machine which violates the second law of thermodynamics. So lets consider adding some friction for realism. Now, experimental scientists[?] will tell us that friction will tend to deccelerate the mass and have magnitude proportional to its velocity (i.e. <math>dx/dt</math>). Our new differential equation, expressing the balancing of the acceleration and the forces, is

<math>\frac{d^2x}{dt^2} = - c \frac{dx}{dt} - x</math>

where <math>c</math> is our coefficient of friction, and <math>c > 0</math>. Again looking for solutions of the form <math>A e^{kt}</math>, we find that

<math>k^2 + c k + 1 = 0.</math>

This is a quadratic equation which we can solve. If <math>c < 2</math> we have complex roots <math>a \pm i b</math>, and the solution (with the above boundary conditions) will look like this:

<math>x(t) = e^{at} \left(\cos bt - \frac{a}{b} \sin bt \right) </math>

(We can show that <math>a < 0</math>)

This is a damped oscillator, and the plot of displacement against time would look something like this:

which does resemble how we'd expect a vibrating spring to behave as friction removed the energy from the system.


See also Laplace transform, eigenvalues, eigenvectors, Vector field, Slope field[?], Integration, Partial derivative, Vector calculus



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