Editing Linear differential equation (section)

==First-order equation with variable coefficients==

The general form of a linear ordinary differential equation of order 1, after dividing out the coefficient of {{math|''y''′(''x'')}}, is: 
<math display="block">y'(x) = f(x) y(x) + g(x).</math>

If the equation is homogeneous, i.e. {{math|1=''g''(''x'') = 0}}, one may rewrite and integrate:
<math display="block">\frac{y'}{y}= f, \qquad \log y = k +F, </math>
where {{mvar|k}} is an arbitrary [[constant of integration]] and <math>F=\textstyle\int f\,dx</math> is any [[antiderivative]] of {{mvar|f}}. Thus, the general solution of the homogeneous equation is
<math display="block">y=ce^F,</math>
where {{math|1=''c'' = ''e''<sup>''k''</sup>}} is an arbitrary constant.

For the general non-homogeneous equation, it is useful to multiply both sides of the equation by the [[multiplicative inverse|reciprocal]] {{math|''e''<sup>−''F''</sup>}} of a solution of the homogeneous equation.<ref>Motivation: In analogy to [[completing the square]] technique we write the equation as {{math|1=''y''′ − ''fy'' = ''g''}}, and try to modify the left side so it becomes a derivative. Specifically, we seek an "integrating factor" {{math|1=''h'' = ''h''(''x'')}} such that multiplying by it makes the left side equal to the derivative of {{math|''hy''}}, namely {{math|1=''hy''′ − ''hfy'' = (''hy'')′}}. This means {{math|1=''h''′ = −''hf''}}, so that {{math|1=''h'' = ''e''<sup>−∫ ''f'' ''dx''</sup> = ''e''<sup>−''F''</sup>}}, as in the text.</ref> This gives 
<math display="block">y'e^{-F}-yfe^{-F}= ge^{-F}.</math>
As {{tmath|1=-fe^{-F} = \tfrac{d}{dx} \left(e^{-F}\right),}} the [[product rule]] allows rewriting the equation as
<math display="block">\frac{d}{dx}\left(ye^{-F}\right)= ge^{-F}.</math>
Thus, the general solution is
<math display="block">y=ce^F + e^F\int ge^{-F}dx,</math>
where {{mvar|c}} is a constant of integration, and {{mvar|F}} is any antiderivative of {{mvar|f}} (changing of antiderivative amounts to change the constant of integration).

=== Example ===
Solving the equation
<math display="block">y'(x) + \frac{y(x)}{x} = 3x.</math>
The associated homogeneous equation <math>y'(x) + \frac{y(x)}{x} = 0</math> gives
<math display="block">\frac{y'}{y}=-\frac{1}{x},</math>
that is 
<math display="block">y=\frac{c}{x}.</math>

Dividing the original equation by one of these solutions gives
<math display="block">xy'+y=3x^2.</math>
That is 
<math display="block">(xy)'=3x^2,</math> 
<math display="block">xy=x^3 +c,</math>
and
<math display="block">y(x)=x^2+c/x.</math>
For the initial condition
<math display="block">y(1)=\alpha,</math>
one gets the particular solution
<math display="block">y(x)=x^2+\frac{\alpha-1}{x}.</math>