LINEAR DIFFERENTIAL EQUATION

In mathematics, a 'linear differential equation' is a differential equation of the form
:''Ly = f'',
where the differential operator ''L'' is a linear operator, ''y'' is the unknown function, and the right hand side ''f'' is a given function. The linearity condition on ''L'' rules out operations such as taking the square of the derivative of ''y''; but permits, for example, taking the second derivative of ''y''. Therefore a fairly general form of such an equation would be
: $a\_n(x)\; D^n\; y(x)\; +\; a\_\{n-1\}(x)D^\{n-1\}\; y(x)\; +\; cdots\; +\; a\_1(x)\; D\; y(x)\; +\; a\_0(x)\; y(x)\; =f(x)$
where ''D'' is the differential operator ''d/dx'' (i.e. ''Dy = y' '', ''D²y = y",... ''), and the ''a_i'' are given functions. Such an equation is said to have 'order' ''n'', the index of the highest derivative of ''f'' that is involved. (Assuming a possibly existing coefficient ''a_n'' of this derivative to be non zero, it is eliminated by dividing through it. In case it can become zero, different cases must be considered separately for the analysis of the equation.)
If ''y'' is assumed to be a function of only one variable, one speaks about an ordinary differential equation, else the derivatives and their coefficients must be understood as (contracted) vectors, matrices or tensors of higher rank, and we have a (linear) partial differential equation.
The case where ''f'' = 0 is called a 'homogeneous equation' and its solutions are called 'complementary functions'. It is particularly important to the solution of the general case, since any complementary function can be added to a solution of the inhomogeneous equation to give another solution (by a method traditionally called ''particular integral and complementary function''). When the ''a_i'' are numbers, the equation is said to have ''constant coefficients''.

Contents

Homogeneous equations with constant coefficients

Examples

Simple harmonic oscillator

Damped harmonic oscillator

Nonhomogeneous equation with constant coefficients

Example

Equation with variable coefficients

Examples

First order equation

Examples

See also

Homogeneous equations with constant coefficients

To solve such an equation one makes a substitution
: $y\; =\; e^\{lambda\; x\}\; !$,
to form the ''characteristic equation''
: $\{lambda^n\; +a\_\{n-1\}lambda^\{n-1\}+cdots+a\_1lambda+a\_0\; =\; 0\}$
to obtain the solutions
: $lambda=s\_0,\; s\_1,\; dots,\; s\_\{n-1\}.$
When this polynomial has distinct roots, we have immediately ''n'' solutions to the differential equation in the form
: $\{y\_i(x)=e^\{s\_i\; x\}.\}$
It can be shown that these are linearly independent, by applying the Vandermonde determinant. Since homogenous linear DEs obey the superposition principle, their linear combinations, with ''n'' coefficients, should provide a complete solution. So it proves: it is known that the general solution to the homogeneous equation can be formed from a linear combination of the ''y''_''i'', ie.,
: $\{y\_H(x)=A\_0\; y\_0(x)+A\_1\; y\_1+cdots+A\_\{n-1\}\; y\_\{n-1\}\}$
Where the solutions are not distinct, it may be necessary to multiply them by some power of ''x'' to obtain linear independence; the general solution therefore involves the product of polynomials, of degrees bounded in terms of the multiplicities of the roots, and exponentials.
The first method of solving linear ordinary differential
equations with constant coefficients is due to Euler, who realized that solutions have the form $e^\{z\; x\}$, for possibly-complex values of $z$. Thus to solve
:
we set $y=e^\{z\; x\}$, leading to
:$z^n\; e^\{zx\}\; +\; A\_1\; z^\{n-1\}\; e^\{zx\}\; +\; cdots\; +\; A\_n\; e^\{zx\}\; =\; 0$
so dividing by $e^\{zx\}$ gives the ''n''th-order polynomial
:$F(z)\; =\; z^\{n\}\; +\; A\_\{1\}z^\{n-1\}\; +\; cdots\; +\; A\_n\; =\; 0$
In short the terms
:
of the original differential equation are replaced by ''z''^''k''. Solving the polynomial gives ''n'' values of ''z'', $z\_1,\; dots,z\_n$. Plugging those values into $e^\{z\_i\; x\}$ gives a basis for the solution; any linear combination of these basis functions will satisfy the differential equation.
This equation ''F''(''z'') = 0, is the "characteristic"
equation considered later by Monge and Cauchy.
{{ExampleSidebar|35%|$y\text{'}\text{'}-2y\text{'}+2y\text{'}\text{'}-2y\text{'}+y=0\; ,$
has the characteristic equation
$z^4-2z^3+2z^2-2z+1=0\; ,$.
This has zeroes, ''i'', −''i'', and 1 (multiplicity 2). The solution basis is then
$e^\{ix\}\; ,,\; e^\{-ix\}\; ,,\; e^x\; ,,\; xe^x\; ,.$
This corresponds to the real-valued solution basis
$cos\; x\; ,,\; sin\; x\; ,,\; e^x\; ,,\; xe^x\; ,.$}}
If ''z'' is a (possibly not real) zero(or root) of ''F''(''z'') of multiplicity ''m'' and $kin\{0,1,dots,m-1\}\; ,$ then $y=x^ke^\{zx\}\; ,$ is a solution of the ODE. These functions make up a basis of the ODE's solutions.
If the ''A_i'' are real then real-valued solutions are preferable. Since the non-real ''z'' values will come in conjugate pairs, so will their corresponding ''y''s; replace each pair with their linear combinations Re(''y'') and Im(''y'').
A case that involves complex roots can be solved with the aid of Euler's formula.

Examples

Given $y\text{'}\text{'}-4y\text{'}+5y=0\; ,$. The characteristic equation is $z^2-4z+5=0\; ,$ which has zeroes 2+''i'' and 2−''i''. Thus the solution basis $\{y\_1,y\_2\}$ is $\{e^\{(2+i)x\},e^\{(2-i)x\}\}\; ,$. Now ''y'' is a solution iff $y=c\_1y\_1+c\_2y\_2\; ,$ for $c\_1,c\_2inmathbb\; C$.
Because the coefficients are real,

★ we are likely not interested in the complex solutions

★ our basis elements are mutual conjugates
The linear combinations
: and
:
will give us a real basis in $\{u\_1,u\_2\}$.

Simple harmonic oscillator

The second order differential equation
:$D^2\; y\; =\; -k^2\; y,$
which represents a simple harmonic oscillator, can be restated as
:$(D^2\; +\; k^2)\; y\; =\; 0.$
The expression in parenthesis can be factored out, yielding
:$(D\; +\; i\; k)\; (D\; -\; i\; k)\; y\; =\; 0,$
which has a pair of linearly independent solutions, one for
:$(D\; -\; i\; k)\; y\; =\; 0$
and another for
:$(D\; +\; i\; k)\; y\; =\; 0.$
The solutions are, respectively,
:$y\_0\; =\; A\_0\; e^\{i\; k\; x\}$
and
:$y\_1\; =\; A\_1\; e^\{-i\; k\; x\}.$
These solutions provide a basis for the two-dimensional "solution space" of the second order differential equation: meaning that linear combinations of these solutions will also be solutions. In particular, the following solutions can be constructed
:$y\_\{0\text{'}\}\; =\; \{A\_0\; e^\{i\; k\; x\}\; +\; A\_0\; e^\{-i\; k\; x\}\; over\; 2\}\; =\; A\_0\; cos\; (k\; x)$
and
:$y\_\{1\text{'}\}\; =\; \{A\_1\; e^\{i\; k\; x\}\; -\; A\_1\; e^\{-i\; k\; x\}\; over\; 2\; i\}\; =\; A\_1\; sin\; (k\; x).$
These last two trigonometric solutions are linearly independent, so they can serve as another basis for the solution space, yielding the following general solution:
:$y\_H\; =\; A\_0\; cos\; (k\; x)\; +\; A\_1\; sin\; (k\; x).$

Damped harmonic oscillator

Given the equation for the damped harmonic oscillator:
:$left(D^2\; +\; \{b\; over\; m\}\; D\; +\; omega\_0^2$
ight) y = 0,
the expression in parentheses can be factored out: first obtain the characteristic equation by replacing ''D'' with λ. This equation must be satisfied for all ''y'', thus:
:$lambda^2\; +\; \{b\; over\; m\}\; lambda\; +\; omega\_0^2\; =\; 0.$
Solve using the quadratic formula:
:$lambda\; =\; \{-b/m\; pm\; sqrt\{b^2\; /\; m^2\; -\; 4\; omega\_0^2\}\; over\; 2\}.$
Use these data to factor out the original differential equation:
:$left(D\; +\; \{b\; over\; 2\; m\}\; -\; sqrt\{\{b^2\; over\; 4\; m^2\}\; -\; omega\_0^2\}$
ight) left(D + {b over 2m} + sqrt{{b^2 over 4 m^2} - omega_0^2}
ight) y = 0.
This implies a pair of solutions, one corresponding to
:$left(D\; +\; \{b\; over\; 2\; m\}\; -\; sqrt\{\{b^2\; over\; 4\; m^2\}\; -\; omega\_0^2\}$
ight) y = 0
and another to
:$left(D\; +\; \{b\; over\; 2m\}\; +\; sqrt\{\{b^2\; over\; 4\; m^2\}\; -\; omega\_0^2\}$
ight) y = 0
The solutions are, respectively,
:$y\_0\; =\; A\_0\; e^\{-omega\; x\; +\; sqrt\{omega^2\; -\; omega\_0^2\}\; x\}\; =\; A\_0\; e^\{-omega\; x\}\; e^\{sqrt\{omega^2\; -\; omega\_0^2\}\; x\}$
and
:$y\_1\; =\; A\_1\; e^\{-omega\; x\; -\; sqrt\{omega^2\; -\; omega\_0^2\}\; x\}\; =\; A\_1\; e^\{-omega\; x\}\; e^\{-sqrt\{omega^2\; -\; omega\_0^2\}\; x\}$
where ω = ''b'' / 2''m''. From this linearly independent pair of solutions can be constructed another linearly independent pair which thus serve as a basis for the two-dimensional solution space:
:$y\_H\; (A\_0,\; A\_1)\; (x)\; =\; left(A\_0\; sinh\; sqrt\{omega^2\; -\; omega\_0^2\}\; x\; +\; A\_1\; cosh\; sqrt\{omega^2\; -\; omega\_0^2\}\; x$
ight) e^{-omega x}.
However, if |ω| < |ω₀| then it is preferable to get rid of the consequential imaginaries, expressing the general solution as
:$y\_H\; (A\_0,\; A\_1)\; (x)\; =\; left(A\_0\; sin\; sqrt\{omega\_0^2\; -\; omega^2\}\; x\; +\; A\_1\; cos\; sqrt\{omega\_0^2\; -\; omega^2\}\; x$
ight) e^{-omega x}.
This latter solution corresponds to the underdamped case, whereas the former one corresponds to the overdamped case: the solutions for the underdamped case oscillate whereas the solutions for the overdamped case do not.

Nonhomogeneous equation with constant coefficients

To obtain the solution to the 'non-homogeneous equation' (sometimes called 'inhomogeneous equation'), find a particular solution ''y''_''P''(''x'') by either the method of undetermined coefficients or the method of variation of parameters; the general solution to the linear differential equation is the sum of the general solution of the related homogeneous equation and the particular solution.
Suppose we face
:
For later convenience, define the characteristic polynomial
:$P(v)=v^n+A\_1v^\{n-1\}+cdots+A\_n.$
We find the solution basis $\{y\_1,y\_2,ldots,y\_n\}$ as in the homogeneous (''f''=0) case. We now seek a 'particular solution' ''y_p'' by the 'variation of parameters' method. Let the coefficients of the linear combination be functions of ''x'':
:$y\_p=u\_1y\_1+u\_2y\_2+cdots+u\_ny\_n.$
Using the "operator" notation $D=d/dx$ and a broad-minded use of notation, the ODE in question is $P(D)y=f$; so
:$f=P(D)y\_p=P(D)(u\_1y\_1)+P(D)(u\_2y\_2)+cdots+P(D)(u\_ny\_n).$
With the constraints
:$0=u\text{'}\_1y\_1+u\text{'}\_2y\_2+cdots+u\text{'}\_ny\_n$
:$0=u\text{'}\_1y\text{'}\_1+u\text{'}\_2y\text{'}\_2+cdots+u\text{'}\_ny\text{'}\_n$
:$cdots$
:$0=u\text{'}\_1y^\{(n-2)\}\_1+u\text{'}\_2y^\{(n-2)\}\_2+cdots+u\text{'}\_ny^\{(n-2)\}\_n$
the parameters commute out, with a little "dirt":
:$f=u\_1P(D)y\_1+u\_2P(D)y\_2+cdots+u\_nP(D)y\_n+u\text{'}\_1y^\{(n-1)\}\_1+u\text{'}\_2y^\{(n-1)\}\_2+cdots+u\text{'}\_ny^\{(n-1)\}\_n.$
But $P(D)y\_j=0$, therefore
:$f=u\text{'}\_1y^\{(n-1)\}\_1+u\text{'}\_2y^\{(n-1)\}\_2+cdots+u\text{'}\_ny^\{(n-1)\}\_n.$
This, with the constraints, gives a linear system in the $u\text{'}\_j$. This much can always be solved; in fact, combining Cramer's rule with the Wronskian,
:
The rest is a matter of integrating $u\text{'}\_j.$
The particular solution is not unique; $y\_p+c\_1y\_1+cdots+c\_ny\_n$ also satisfies the ODE for any set of constants ''c_j''.

Example

Suppose $y\text{'}\text{'}-4y\text{'}+5y=sin(kx)$. We take the solution basis found above $\{e^\{(2+i)x\},e^\{(2-i)x\}\}$.
:{|
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|$W,$
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|$=-2ie^\{4x\},$
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:{|
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|$u\text{'}\_1,$
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:{|
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|$u\text{'}\_2,$
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Using the list of integrals of exponential functions
:{|
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|$u\_1,$
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ight)
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:{|
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|$u\_2,$
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ight).
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And so
:{|
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|$y\_p,$
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ight)
+rac{i}{2(3-4i+k^2)}left((i-2)sin(kx)-kcos(kx)
ight)
|-
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(Notice that ''u''₁ and ''u''₂ had factors that canceled ''y''₁ and ''y''₂; that is typical.)
For interest's sake, this ODE has a physical interpretation as a driven damped harmonic oscillator; ''y_p'' represents the steady state, and $c\_1y\_1+c\_2y\_2$ is the transient.

Equation with variable coefficients

A linear ODE of order ''n'' with variable coefficients has the general form
:$p\_\{n\}(x)y^\{(n)\}(x)\; +\; p\_\{n-1\}(x)\; y^\{(n-1)\}(x)\; +\; ldots\; +\; p\_0(x)\; y(x)\; =\; r(x).$

Examples

A particular simple example is the Cauchy-Euler equation often used in engineering
:$x^n\; y^\{(n)\}(x)\; +\; a\_\{n-1\}\; x^\{n-1\}\; y^\{(n-1)\}(x)\; +\; ldots\; +\; a\_0\; y(x)\; =\; 0.$

First order equation

{{ExampleSidebar|35%|$-3y\text{'}\text{'}+4y\text{'}=5\; ,$
with the initial condition
$fleft(0$
ight)=2 ,.
Using the general solution method:
$f=e^\{-3x\}left(int\; 2\; e^\{3x\}\; dx\; +\; kappa$
ight) ,.
The integration is done from 0 to x, giving:
$f=e^\{-3x\}left(2/3left(\; e^\{3x\}-e^0$
ight) + kappa
ight) ,.
Then we can reduce to:
$f=2/3left(1-e^\{-3x\}$
ight) + e^{-3x}kappa ,.
Assume that kappa is 2 from the initial condition.}}
For a first-order linear ODE, with coefficients that may or may not vary with ''x'':
$y\text{'}(x)\; +\; p(x)\; y(x)\; =\; r(x)$
Then,
:$y=e^\{-a(x)\}left(int\; r(x)\; e^\{a(x)\}\; dx\; +\; kappa$
ight)
where $kappa$ is the constant of integration, and
$a(x)=int\{p(x)dx\}.$
This proof comes from Jean Bernoulli. Let
:$y^prime\; +\; py\; =\; r$
Suppose for some unknown functions ''u''(''x'') and ''v''(''x'') that ''y'' = ''uv''.
Then
:$y^prime\; =\; u^prime\; v\; +\; u\; v^prime$
Substituting into the differential equation,
:$u^prime\; v\; +\; u\; v^prime\; +\; puv\; =\; r$
Now, the most important step: Since the differential equation is ''linear'' we can split this into two independent equations and write
:$u^prime\; v\; +\; puv\; =\; 0$
:$u\; v^prime\; =\; r$
Since v is not zero, the top equation becomes
:$u^prime\; +\; pu\; =\; 0$
The solution of this is
:$u\; =\; e^\{\; -\; int\; p\; dx\; \}$
Substituting into the second equation
:$v\; =\; int\; r\; e^\{\; int\; p\; dx\}\; +\; C$
Since ''y'' = ''uv'', for arbitrary constant ''C''
:$y\; =e^\{\; -\; int\; p\; dx\; \}\; left(\; int\; r\; e^\{\; int\; p\; dx\; \}\; +\; C$
ight)

Examples

Consider a first order differential equation with constant coefficients:
:
This equation is particularly relevant to first order systems such as RC circuits and mass-damper systems.
In this case, ''p''(''x'') = b, ''r''(''x'') = 1.
Hence its solution is
:$y(x)\; =\; e^\{-bx\}\; left(\; e^\{bx\}/b+\; C$
ight) = 1/b + C e^{-bx} .

See also

★ Laplace transform

★ Fourier transform

★ Differential equation