Examples of differential equations: Difference between revisions

Revision as of 16:11, 1 May 2014

Revision as of 11:20, 5 May 2014

Line 18:

:<math>\frac{dy}{y} = -f(t)\, 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.

We solve the transformed equation with the variables already separated by [[Integral Calculus|Integrating]],

:<math>\ln |y| = \left(-\int f(t)\,dt\right) + C\,</math>

Line 36:

Some elaboration is needed because ''&fnof;''(''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.

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{dy}{dt} + \alpha y = 0, y(1)=2, y(2)=1</math>

Line 43:

==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{dy}{dx} + p(x)y = q(x)</math>

The method for solving this equation relies on a special integrating factor, ''&mu;'':

:<math>\mu = e^{\int p(x)\, dx}</math>

Line 90:

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

See a [http://www.wolframalpha.com/input/?i=x%27%27%3D-x solution] by [[WolframAlpha]].

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&nbsp;''t''&nbsp;=&nbsp;0).

Line 146:

can be easily symbolically

[http://www.wolframalpha.com/input/?i=y%27+%3D+{{1%2C2}}%2C{{2%2C-2}}.y%2B+{t%2C+sin%28t%29} solved]

in [[WolframAlpha]].