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#REDIRECT [[Differential equation#Examples]]
'''[[Differential equation]]s''' arise in many problems in [[physics]], [[engineering]], and other sciences. The following examples show how to solve differential equations in a few simple cases when an exact solution exists.
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== Separable first-order ordinary differential equations ==
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{{see also|Separable partial differential equation}}
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Equations in the form {{math|1=''{{sfrac|dy|dx}}''= ''f'' (''x'')''g''(''y'')}} are called separable and solved by {{math|1={{sfrac|''dy''|''g''(''y'')}} = ''f'' (''x'') ''dx''}} and thus
}}
:<math>\int\frac {dy}{g(y)} = \int f(x) \, dx.</math>
Prior to dividing by {{math|''g''(''y'')}}, one needs to check if there are stationary (also called equilibrium) solutions with constant {{mvar|y}} satisfying {{math|1=''g''(''y'') = 0}}.
==Separable (homogeneous) first-order linear ordinary differential equations==
{{see also|Separable partial differential equation}}
A separable ''linear'' [[ordinary differential equation]] of the first order must be homogeneous and has the general form
:<math>\frac{dy}{dt} + f(t) y = 0</math>
where {{math|''f'' (''t'')}} is some known [[function (mathematics)|function]]. We may solve this by [[separation of variables]] (moving the {{mvar|y}} terms to one side and the {{mvar|t}} terms to the other side),
:<math>\frac{dy}{y} = -f(t)\, dt</math>
Since the separation of variables in this case involves dividing by {{mvar|y}}, we must check if the constant function {{math|1=''y'' = 0}} is a solution of the original equation. Trivially, if {{math|1=''y'' = 0}} then {{math|1=''y''′ = 0}}, so {{math|1=''y'' = 0}} is actually a solution of the original equation. We note that {{math|1=''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>
where {{mvar|C}} is an arbitrary constant. Then, by [[exponentiation]], we obtain
:<math>y = \pm e^{\left(-\int f(t)\,dt\right) + C} = \pm e^C e^{-\int f(t)\,dt}.</math>
Here, {{math|''e''<sup>''C''</sup> > 0}}, so <math>\pm e^C \neq 0</math>. But we have independently checked that {{math|1=''y'' = 0}} is also a solution of the original equation, thus
:<math>y = A e^{-\int f(t)\,dt}.</math>
with an arbitrary constant {{mvar|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{dy}{dt} + f(t) y = -f(t) \cdot A e^{-\int f(t)\,dt} + f(t) \cdot A e^{-\int f(t)\,dt} = 0</math>
Some elaboration is needed because {{math|''f'' (''t'')}} may 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|1=''f''(''t'') = ''α''}} is a constant, the solution is particularly simple, <math>y = A e^{-\alpha t}</math> and describes, e.g., if {{math|''α'' > 0}}, the exponential decay of radioactive material at the macroscopic level. If the value of {{mvar|α}} 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>
gives {{math|1=''α'' = ln(2)}} 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{dy}{dx} + p(x)y = q(x)</math>
The method for solving this equation relies on a special integrating factor, {{math|1=''μ'' = ''μ''(''x'')}}:<!--signify integrating factors should only be dependent on ''x''-->
:<math>\mu = e^{\int_{x_0}^x p(t)\, 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{d\mu}{dx} = e^{\int_{x_0}^x p(t)\, dt} \ \underbrace{\frac{d}{dx} \int_{x_0}^x p(t)\, dt}_{p(x)} = \mu p(x)</math>
Multiply both sides of the original differential equation by {{mvar|μ}} to get:
:<math>\mu \frac{dy}{dx} + \mu p(x)y = \mu q(x)</math>
Because of the special {{mvar|μ}} we picked, we may substitute {{math|{{sfrac|''dμ''|''dx''}}}} for {{math|''μ p''(''x'')}}, simplifying the equation to:
:<math>\mu{\frac{dy}{dx}} + y \frac{d\mu}{dx} = \mu q(x)</math>
Using the [[product rule (calculus)|product rule]] in reverse, we get:
:<math>\frac{d}{dx}{(\mu y)} = \mu q(x)</math>
Integrating both sides with respect to {{mvar|x}}:
:<math>\mu y = C + \int\mu q(x)\, dx</math>
Finally, to solve for {{mvar|y}} we divide both sides by {{mvar|μ}}:
:<math>y = \frac{C + \int\mu q(x)\, dx}{\mu}</math>
Since {{mvar|μ}} is a function of {{mvar|x}}, we cannot simplify any further directly.
==Second-order linear ordinary differential equations==
===A simple example===
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 {{mvar|t}} as {{math|''x''(''t'')}}. Now, using [[Newton's laws of motion|Newton's second law]] we can write (using convenient units):
: <math>m\frac{d^2x}{dt^2} + kx = 0,</math>
where {{mvar|m}} is the mass and {{mvar|k}} is the spring constant that represents a measure of spring stiffness. For simplicity's sake, let us take {{math|1=''m'' = ''k''}} as an example.
If we look for solutions that have the form {{math|''Ce''<sup>λ''t''</sup>}}, where {{mvar|C}} is a constant, we discover the relationship {{math|1=λ<sup>2</sup> + 1 = 0}}, and thus {{math|λ}} must be one of the [[complex number]]s {{mvar|i}} or {{math|−''i''}}. Thus, using [[Euler's formula]] we can say that the solution must be of the form:
: <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 {{mvar|A}} and {{mvar|B}}, we need ''initial conditions'', i.e. equalities that specify the state of the system at a given time (usually {{math|1=''t'' = 0}}).
For example, if we suppose at {{math|1=''t'' = 0}} the extension is a unit distance ({{math|1=''x'' = 1}}), and the particle is not moving ({{math|1=''x''′ = 0}}). We have
: <math>x(0) = A \cos 0 + B \sin 0 = A = 1,</math>
and so {{math|1=''A'' = 1}}.
: <math>x'(0) = -A \sin 0 + B \cos 0 = B = 0,</math>
and so {{math|1=''B'' = 0}}.
Therefore {{math|1=''x''(''t'') = cos ''t''}}. This is an example of [[simple harmonic motion]].
See a [http://www.wolframalpha.com/input/?i=x%27%27%3D-x%2Cx%280%29%3D1%2Cx%27%280%29%3D0 solution] by [[Wolfram Alpha]].
===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. {{math|''dx''/''dt''}}). Our new differential equation, expressing the balancing of the acceleration and the forces, is
: <math>m\frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0,</math>
where {{mvar|c}} is the damping coefficient representing friction. Again looking for solutions of the form {{math|''Ce''<sup>λ''t''</sup>}}, we find that
: <math>m\lambda^2 + c \lambda + k = 0.</math>
This is a [[quadratic equation]] which we can solve. If {{math|''c''{{sup|2}} < 4''km''}} there are two complex conjugate roots {{math|''a'' ± ''bi''}}, 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>
Let us for simplicity take {{math|1=''m'' = 1}}, then {{math|1=0 < ''c'' = −2''a''}}, and {{math|1=''k'' = ''a''{{sup|2}} + ''b''{{sup|2}}}}.
The equation can be also solved in MATLAB symbolic toolbox as
<syntaxhighlight lang="matlab">
x = dsolve('D2x+c*Dx+k*x=0','x(0)=1','Dx(0)=0')
</syntaxhighlight>
although the solution looks rather ugly,
<syntaxhighlight lang="matlab">
x = (c + (c^2 - 4*k)^(1/2))/(2*exp(t*(c/2 - (c^2 - 4*k)^(1/2)/2))*(c^2 - 4*k)^(1/2)) -
(c - (c^2 - 4*k)^(1/2))/(2*exp(t*(c/2 + (c^2 - 4*k)^(1/2)/2))*(c^2 - 4*k)^(1/2))
</syntaxhighlight>
This is a model of a [[Damping|damped oscillator]]. The plot of displacement against time would look like this:
: [[Image:Damped Oscillation2.svg|400px|center]]
which resembles how one would expect a vibrating spring to behave as friction removes energy from the system.
==Linear systems of ODEs==
The following example of a first order linear systems of ODEs
: <math>\begin{align}
y_1' &= y_1 + 2y_2 + t\\
y_2' &= 2y_1 - 2y_2 + \sin(t)
\end{align}</math>
can be easily solved symbolically using [[List_of_numerical_analysis_software|numerical analysis software]].
==See also==
* [[Closed and exact differential forms]]
* [[Ordinary differential equation]]
* [[Bernoulli differential equation]]
== Bibliography ==
* A. D. Polyanin and V. F. Zaitsev, ''Handbook of Exact Solutions for Ordinary Differential Equations'', 2nd Edition, [[Chapman & Hall]]/[[CRC Press]], Boca Raton, 2003; {{ISBN|1-58488-297-2}}.
==External links==
* [http://eqworld.ipmnet.ru/en/solutions/ode.htm Ordinary Differential Equations] at EqWorld: The World of Mathematical Equations.
{{Differential equations topics}}
[[Category:Ordinary differential equations]]
[[Category:Mathematical examples|Differential equations]]
[[Category:Articles with example MATLAB/Octave code]]
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