Content deleted Content added
move to footer |
m upright 'd' in 'dx/dt' and similar |
||
Line 3:
== Separable first-order ordinary differential equations ==
{{see also|Separable partial differential equation}}
Equations in the form <math>\frac {
<math>\int\frac {\mathrm dy}{g(y)} = \int f(x)\mathrm dx</math>. Prior to dividing by <math>g(y)</math>, one needs to check if there are stationary (also called equilibrium)
solutions <math>y=const</math> satisfying <math>g(y)=0</math>.
Line 12:
must be homogeneous and has the general form
:<math>\frac{\mathrm dy}{\mathrm dt} + f(t) y = 0</math>
where <math>f(t)</math> is some known [[function (mathematics)|function]]. We may solve this by [[separation of variables]] (moving the ''y'' terms to one side and the ''t'' terms to the other side),
:<math>\frac{\mathrm dy}{y} = -f(t)\, \mathrm dt</math>
Since the separation of variables in this case involves dividing by ''y'', we must check if the constant function ''y=0'' is a solution of the original equation. Trivially, if ''y=0'' then ''y'=0'', so ''y=0'' is actually a solution of the original equation. We note that ''y=0'' is not allowed in the transformed equation.
Line 22:
We solve the transformed equation with the variables already separated by [[Integral Calculus|Integrating]],
:<math>\ln |y| = \left(-\int f(t)\,\mathrm dt\right) + C\,</math>
where ''C'' is an arbitrary constant. Then, by [[exponentiation]], we obtain
:<math>y = \pm e^{\left(-\int f(t)\,\mathrm dt\right) + C} = \pm e^{C} e^{-\int f(t)\,\mathrm dt}</math>.
Here, <math>e^{C}>0</math>, so <math>\pm e^{C}\neq 0</math>. But we have independently checked that ''y=0'' is also a solution of the original equation, thus
:<math>y = A e^{-\int f(t)\,\mathrm dt}</math>.
with an arbitrary constant ''A'', which covers all the cases. It is easy to confirm that this is a solution by plugging it into the original differential equation:
:<math>\frac{\mathrm dy}{\mathrm dt} + f(t) y = -f(t) \cdot A e^{-\int f(t)\,\mathrm dt} + f(t) \cdot A e^{-\int f(t)\,\mathrm dt} = 0</math>
Some elaboration is needed because ''ƒ''(''t'') might not even be integrable. One must also assume something about the domains of the functions involved before the equation is fully defined. The solution above assumes the [[real number|real]] case.
Line 38:
If <math>f(t)=\alpha</math> is a constant, the solution is particularly simple, <math>y = A e^{-\alpha t}</math> and describes, e.g., if <math>\alpha>0</math>, the exponential decay of radioactive material at the macroscopic level. If the value of <math>\alpha</math> is not known a priori, it can be determined from two measurements of the solution. For example,
:<math>\frac{\mathrm dy}{\mathrm dt} + \alpha y = 0, y(1)=2, y(2)=1</math>
gives <math>\alpha = \ln(2)</math> and <math>y = 4 e^{-\ln(2) t}= 2^{2-t}</math>.
==Non-separable (non-homogeneous) first-order linear ordinary differential equations==
First-order linear non-homogeneous ODEs (ordinary [[differential equation]]s) are not separable. They can be solved by the following approach, known as an ''[[integrating factor]]'' method. Consider first-order linear ODEs of the general form:
:<math>\frac{\mathrm dy}{\mathrm dx} + p(x)y = q(x)</math>
The method for solving this equation relies on a special integrating factor, ''μ'':
:<math>\mu = e^{\int_{x_0}^x p(t)\, \mathrm dt}</math>
We choose this integrating factor because it has the special property that its derivative is itself times the function we are integrating, that is:
:<math>\frac{\mathrm d
Multiply both sides of the original differential equation by ''μ'' to get:
:<math>\mu{\frac{\mathrm dy}{\mathrm dx}} + \mu{p(x)y} = \mu{q(x)}</math>
Because of the special ''μ'' we picked, we may substitute d''
:<math>\mu{\frac{\mathrm dy}{\mathrm dx}} + y{\frac{\mathrm d
Using the [[product rule (calculus)|product rule]] in reverse, we get:
:<math>\frac{\mathrm d}{\mathrm dx}{(\mu{y})} = \mu{q(x)}</math>
Integrating both sides:
:<math>\mu{y} = \left(\int\mu q(x)\, \mathrm dx\right) + C</math>
Finally, to solve for ''y'' we divide both sides by <math>\mu</math>:
:<math>y = \frac{\left(\int\mu q(x)\, \mathrm dx\right) + C}{\mu}</math>
Since ''μ'' is a function of ''x'', we cannot simplify any further directly.
Line 82:
Suppose a mass is attached to a spring which exerts an attractive force on the mass [[Proportionality (mathematics)|proportional]] to the extension/compression of the spring. For now, we may ignore any other forces ([[gravity]], [[friction]], etc.). We shall write the extension of the spring at a time ''t'' as ''x''(''t''). Now, using [[Newton's laws of motion|Newton's second law]] we can write (using convenient units):
: <math>m\frac{\mathrm d^2x}{\mathrm dt^2} +kx=0,</math>
where ''m'' is the mass and ''k'' is the spring constant that represents a measure of spring stiffness. For simplicity's sake, let us take ''m=k'' as an example.
Line 94:
To determine the unknown constants ''A'' and ''B'', we need ''initial conditions'', i.e. equalities that specify the state of the system at a given time (usually ''t'' = 0).
For example, if we suppose at ''t'' = 0 the extension is a unit distance (''x'' = 1), and the particle is not moving (d''
: <math>x(0) = A \cos 0 + B \sin 0 = A = 1, \, </math>
Line 109:
===A more complicated model===
The above model of an oscillating mass on a spring is plausible but not very realistic: in practice, [[friction]] will tend to decelerate the mass and have magnitude proportional to its velocity (i.e. d''
: <math>m\frac{\mathrm d^2x}{\mathrm dt^2} + c \frac{\mathrm dx}{\mathrm dt} + kx=0,</math>
where <math>c</math> is the damping coefficient representing friction. Again looking for solutions of the form <math>Ce^{\lambda t}</math>, we find that
| |||