Editing Linear differential equation (section)

==Homogeneous equation with constant coefficients==
A homogeneous linear differential equation has ''constant coefficients'' if it has the form
<math display="block">a_0y + a_1y' + a_2y'' + \cdots + a_n y^{(n)} = 0</math>
where {{math|''a''<sub>1</sub>, ..., ''a''<sub>''n''</sub>}} are (real or complex) numbers. In other words, it has constant coefficients if it is defined by a linear operator with constant coefficients.

The study of these differential equations with constant coefficients dates back to [[Leonhard Euler]], who introduced the [[exponential function]] {{math|''e''<sup>''x''</sup>}}, which is the unique solution of the equation {{math|1=''f''′ = ''f''}} such that {{math|1=''f''(0) = 1}}. It follows that the {{mvar|n}}th derivative of {{math|''e''<sup>''cx''</sup> }} is {{math|''c''<sup>''n''</sup>''e''<sup>''cx''</sup>}}, and this allows solving homogeneous linear differential equations rather easily.

Let 
<math display="block">a_0y + a_1y' + a_2y'' + \cdots + a_ny^{(n)} = 0</math>
be a homogeneous linear differential equation with constant coefficients (that is {{math|''a''<sub>0</sub>, ..., ''a''<sub>''n''</sub>}} are real or complex numbers).

Searching solutions of this equation that have the form {{math|''e''<sup>''αx''</sup>}} is equivalent to searching the constants {{mvar|α}} such that
<math display="block">a_0e^{\alpha x} + a_1\alpha e^{\alpha x} + a_2\alpha^2 e^{\alpha x}+\cdots + a_n\alpha^n e^{\alpha x} = 0.</math>
Factoring out {{math|''e''<sup>''αx''</sup>}} (which is never zero), shows that {{mvar|α}} must be a root of the ''characteristic polynomial''
<math display="block">a_0 + a_1t + a_2 t^2 + \cdots + a_nt^n</math>
of the differential equation, which is the left-hand side of the [[Characteristic equation (calculus)|characteristic equation]]
<math display="block">a_0 + a_1t + a_2 t^2 + \cdots + a_nt^n = 0.</math>

When these roots are all [[distinct roots|distinct]], one has {{mvar|n}} distinct solutions that are not necessarily real, even if the coefficients of the equation are real. These solutions can be shown to be [[linearly independent]], by considering the [[Vandermonde determinant]] of the values of these solutions at {{math|1=''x'' = 0, ..., ''n'' – 1}}. Together they form a [[Basis (linear algebra)|basis]] of the [[vector space]] of solutions of the differential equation (that is, the kernel of the differential operator).
{| class="toccolours floatright" style="width:35%; margin: 0.5em 0 0.5em 1em;"
! style="background:#ffffaa; padding: 3px 5px 3px 5px; font-size:larger;" | Example
|-
| style="font-size:100%; padding:0 5px 0 5px;" | 
<math display="block">y''''-2y'''+2y''-2y'+y=0</math>
has the characteristic equation
<math display="block">z^4-2z^3+2z^2-2z+1=0.</math>
This has zeros, {{mvar|i}}, {{math|−''i''}}, and {{math|1}} (multiplicity 2). The solution basis is thus
<math display="block">e^{ix},\; e^{-ix},\; e^x,\; xe^x.</math>
A real basis of solution is thus
<math display="block">\cos x,\; \sin x,\; e^x,\; xe^x.</math>
 
|}
In the case where the characteristic polynomial has only [[simple root]]s, the preceding provides a complete basis of the solutions vector space. In the case of [[multiple root]]s, more linearly independent solutions are needed for having a basis. These have the form
<math display="block">x^ke^{\alpha x},</math>
where {{mvar|k}} is a nonnegative integer, {{mvar|α}} is a root of the characteristic polynomial of multiplicity {{mvar|m}}, and {{math|''k'' < ''m''}}. For proving that these functions are solutions, one may remark that if {{mvar|α}} is a root of the characteristic polynomial of multiplicity {{mvar|m}}, the characteristic polynomial may be factored as {{math|''P''(''t'')(''t'' − ''α'')<sup>''m''</sup>}}. Thus, applying the differential operator of the equation is equivalent with applying first {{mvar|m}} times the operator {{nowrap|<math display="inline"> \frac{d}{dx} - \alpha </math>,}} and then the operator that has {{mvar|P}} as characteristic polynomial. By the [[Shift theorem|exponential shift theorem]],
<math display="block">\left(\frac{d}{dx}-\alpha\right)\left(x^ke^{\alpha x}\right)= kx^{k-1}e^{\alpha x},</math>

and thus one gets zero after {{math|''k'' + 1}} application of {{nowrap|1=<math display="inline"> \frac{d}{dx} - \alpha </math>.}}

As, by the [[fundamental theorem of algebra]], the sum of the multiplicities of the roots of a polynomial equals the degree of the polynomial, the number of above solutions equals the order of the differential equation, and these solutions form a basis of the vector space of the solutions.

In the common case where the coefficients of the equation are real, it is generally more convenient to have a basis of the solutions consisting of [[real-valued function]]s. Such a basis may be obtained from the preceding basis by remarking that, if {{math|''a'' + ''ib''}} is a root of the characteristic polynomial, then {{math|''a'' – ''ib''}} is also a root, of the same multiplicity. Thus a real basis is obtained by using [[Euler's formula]], and replacing <math>x^ke^{(a+ib)x}</math> and <math>x^ke^{(a-ib)x}</math> by <math>x^ke^{ax} \cos(bx)</math> and <math>x^ke^{ax} \sin(bx)</math>. 

===Second-order case===
A homogeneous linear differential equation of the second order may be written 
<math display="block">y'' + ay' + by = 0,</math>
and its characteristic polynomial is
<math display="block">r^2 + ar + b.</math>

If {{mvar|a}} and {{mvar|b}} are [[real number|real]], there are three cases for the solutions, depending on the discriminant {{math|1=''D'' = ''a''<sup>2</sup> − 4''b''}}. In all three cases, the general solution depends on two arbitrary constants {{math|''c''<sub>1</sub>}} and {{math|''c''<sub>2</sub>}}.

* If {{math|''D'' > 0}}, the characteristic polynomial has two distinct real roots {{mvar|α}}, and {{mvar|β}}. In this case, the general solution is <math display="block">c_1 e^{\alpha x} + c_2 e^{\beta x}.</math>
* If {{math|1=''D'' = 0}}, the characteristic polynomial has a double root {{math|−''a''/2}}, and the general solution is <math display="block">(c_1 + c_2 x) e^{-ax/2}.</math>
* If {{math|''D'' < 0}}, the characteristic polynomial has two [[complex conjugate]] roots {{math|''α'' ± ''βi''}}, and the general solution is <math display="block">c_1 e^{(\alpha + \beta i)x} + c_2 e^{(\alpha - \beta i)x},</math> which may be rewritten in real terms, using [[Euler's formula]] as <math display="block"> e^{\alpha x} (c_1\cos(\beta x) + c_2 \sin(\beta x)).</math>

Finding the solution {{math|''y''(''x'')}} satisfying {{math|1=''y''(0) = ''d''<sub>1</sub>}} and {{math|1=''y''′(0) = ''d''<sub>2</sub>}}, one equates the values of the above general solution at {{math|0}} and its derivative there to {{math|''d''<sub>1</sub>}} and {{math|''d''<sub>2</sub>}}, respectively. This results in a linear system of two linear equations in the two unknowns {{math|''c''<sub>1</sub>}} and {{math|''c''<sub>2</sub>}}. Solving this system gives the solution for a so-called [[Cauchy boundary condition|Cauchy problem]], in which the values at {{math|0}} for the solution of the DEQ and its derivative are specified.