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== Separable first-order ordinary differential equations ==
{{see also|Separable partial differential equation}}
Equations in the form {{<math|1=''>\frac {dy}{sfrac|dy|dx}}'' = ''f'' (''x'')''g''(''y'')}}</math> are called separable and solved by {{<math|1=>\frac {{sfrac|''dy''|''}{g''(''y'')}} = ''f'' (''x'') ''dx''}}</math> and thus
:<math>\int\frac {dy}{g(y)} = \int f(x) \, 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>.
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'')}}</math> 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'' = 0}} is a solution of the original equation. Trivially, if {{math|1=''y=0'' = 0}} then {{math|1=''y''′ = 0}}'', so {{math|1=''y=0'' = 0}} is actually a solution of the original equation. We note that {{math|1=''y=0'' = 0}} is not allowed in the transformed equation.
We solve the transformed equation with the variables already separated by [[Integral Calculus|integratingIntegrating]],
:<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}>''C''0</supmath> > 0}}, so {{<math|±''>\pm e''<sup>''^{C''}\neq 0</supmath> ≠ 0}}. But we have independently checked that {{math|1=''y=0'' = 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''&thinspfnof;''(''t'')}} may 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|1=''>f''(''t'') = ''α''}}\alpha</math> is a constant, the solution is particularly simple, <math>y = A e^{-\alpha t}</math> and describes, e.g., if {{<math|''α'' > \alpha>0}}</math>, the exponential decay of radioactive material at the macroscopic level. If the value of {{mvar|α}}<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>
gives {{<math|1=''α''>\alpha = \ln(2)}}</math> and {{<math|1=''>y'' = 4'' e''<sup>−ln^{-\ln(2)'' t''</sup> }= 2<sup>2−''^{2-t''}</supmath>}}.
==Non-separable (non-homogeneous) first-order linear ordinary differential equations==
:<math>\frac{dy}{dx} + p(x)y = q(x)</math>
The method for solving this equation relies on a special integrating factor, {{math|1=''μμ'' = ''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}^xcdot p(t)\, dt}_{p(x)} = \mu p(x)</math>
Multiply both sides of the original differential equation by {{mvar|μ}}''μ'' to get:
:<math>\mu {\frac{dy}{dx}} + \mu {p(x)y} = \mu {q(x)}</math>
Because of the special {{mvar|μ}}''μ'' we picked, we may substitute {{math|{{sfrac|''dμdμ''|/''dx''}}}} for {{math|''μ&hairspmu;'' ''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 + \left(\int\mu q(x)\, dx\right) + C</math>
Finally, to solve for {{mvar|''y}}'' we divide both sides by {{mvar|μ}}<math>\mu</math>:
:<math>y = \frac{C + \left(\int\mu q(x)\, dx\right) + C}{\mu}</math>
Since {{mvar|μ}}''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''^{\lambda t''}</supmath>}}, where {{mvar|''C}}'' is a constant, we discover the relationship {{<math|1=λ<sup>\lambda^2</sup> + 1 = 0}}</math>, and thus {{<math|λ}}>\lambda</math> must be one of the [[complex number]]s {{mvar|<math>i}}</math> or {{<math|−''>-i''}}</math>. 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=''xdx''′ /''dt'' = 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|<math>c}}</math> is the damping coefficient representing friction. Again looking for solutions of the form {{math|''Ce''<supmath>λ''Ce^{\lambda t''}</supmath>}}, 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''}}4km</math> there are two complex conjugate roots {{math|''a'' ± ''biib''}}, 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}}</math>, then {{<math|1=>0 < ''c'' = −2''a''}},-2a</math> and {{<math|1=''>k'' = ''a''{{sup|^2}} + ''b''{{sup|^2}}}}</math>.
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,
The following example of a first order linear systems of ODEs
: <math>\begin{align} y_1'=y_1+2y_2+t</math>
: <math> y_2' &= 2y_1 - 2y_2 + \sin(t) </math>▼
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]].
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