Linear differential equation: Definition from Answers.com

In mathematics, a linear differential equation is of the form

$Ly = f \,$

where the differential operator L is a linear operator, y is the unknown function (such as a function of time y(t)), and the right hand side ƒ is a given function of the same nature as y (called the source term). For a function dependent on time we may write the equation more expressively as

$L y(t) = f(t) \,$

and, even more precisely by bracketing

$L [y(t)] = f(t) \,$

The linear operator L may be considered to be of the form^[1]

$L_n(y) \equiv \frac{d^n y}{dt^n} + A_1(t)\frac{d^{n-1}y}{dt^{n-1}} + \cdots + A_{n-1}(t)\frac{dy}{dt} + A_n(t)y \,$

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. It is convenient to rewrite this equation in an operator form

$L_n(y) \equiv \left[\,D^n + A_{1}(t)D^{n-1} + \cdots + A_{n-1}(t) D + A_n(t)\right] y$

where D is the differential operator d/dt (i.e. Dy = y' , D²y = y",... ), and the A_n are given functions.

Such an equation is said to have order n, the index of the highest derivative of y that is involved.

A typical simple example is the linear differential equation used to model radioactive decay^[2]. Let N(t) denote the number of radioactive atoms in some sample of material (such as a portion of the cloth of the Shroud of Turin^[3]) at time t. Then for some constant k > 0, the number of radioactive atoms which decay can be modelled by

$\frac{dN}{dt} = -k N\,$

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 ƒ = 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.

1 Homogeneous equations with constant coefficients
- 1.1 Examples
  - 1.1.1 Simple harmonic oscillator
  - 1.1.2 Damped harmonic oscillator
2 Nonhomogeneous equation with constant coefficients
- 2.1 Example
3 Equation with variable coefficients
- 3.1 Examples
4 First order equation
- 4.1 Examples
5 See also
6 Notes
7 References

Homogeneous equations with constant coefficients

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 $z$ . The exponential function is one of the few functions that keep its shape even after differentiation. In order for the sum of multiple derivatives of a function to sum up to zero, the derivatives must cancel each other out and the only way for them to do so is for the derivatives to have the same form as the initial function. Thus, to solve

$y^{(n)} + A_{1}y^{(n-1)} + \cdots + A_{n}y = 0$

we set $y = e z x$ , leading to

$z^n e^{zx} + A_1 z^{n-1} e^{zx} + \cdots + A_n e^{zx} = 0.$

Division by e^zx gives the nth-order polynomial

$F(z) = z^{n} + A_{1}z^{n-1} + \cdots + A_n = 0.\,$

This equation F(z) = 0, is the "characteristic" equation considered later by Monge and Cauchy.

Formally, the terms

$y^{(k)}\quad\quad(k = 1, 2, \dots, n).$

of the original differential equation are replaced by z^k. Solving the polynomial gives n values of z, z₁, ..., z_n. Substitution of any of those values for z into e^zx gives a solution e^z_ix. Since homogeneous linear differential equations obey the superposition principle, any linear combination of these functions also satisfies the differential equation.

When these roots are all distinct, we have n distinct solutions to the differential equation. It can be shown that these are linearly independent, by applying the Vandermonde determinant, and together they form a basis of the space of all solutions of the differential equation.

Examples

$y''''-2y'''+2y''-2y'+y=0 \,$

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

$\cos x ,\, \sin x ,\, e^x ,\, xe^x \,.$

The preceding gave a solution for the case when all zeros are distinct, that is, each has multiplicity 1. For the general case, if z is a (possibly complex) zero (or root) of F(z) having multiplicity m, then, for $k\in\{0,1,\dots,m-1\} \,$ , $y=x^ke^{zx} \,$ is a solution of the ODE. Applying this to all roots gives a collection of n distinct and linearly independent functions, where n is the degree of F(z). As before, these functions make up a basis of the solution space.

If the coefficients A_i of the differential equation are real, then real-valued solutions are generally preferable. Since non-real roots z then come in conjugate pairs, so do their corresponding basis functions x^ke^zx, and the desired result is obtained by replacing each pair with their real-valued linear combinations Re(y) and Im(y), where y is one of the pair.

A case that involves complex roots can be solved with the aid of Euler's formula.

Examples

Given $y''-4y'+5y=0 \,$ . 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 if and only if $y=c_1y_1+c_2y_2 \,$ for $c_1,c_2\in\mathbb C$ .

Because the coefficients are real,

we are likely not interested in the complex solutions
our basis elements are mutual conjugates

The linear combinations

$u_1=\mbox{Re}(y_1)=\frac{y_1+y_2}{2}=e^{2x}\cos(x) \,$ and

$u_2=\mbox{Im}(y_1)=\frac{y_1-y_2}{2i}=e^{2x}\sin(x) \,$

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

$y_{0'} = {A_0 e^{i k x} + A_1 e^{-i k x} \over 2} = C_0 \cos (k x)$

and

$y_{1'} = {A_0 e^{i k x} - A_1 e^{-i k x} \over 2 i} = C_1 \sin (k x).$

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 = C 0 cos(k x) + C 1 sin(k x).

Damped harmonic oscillator

Given the equation for the damped harmonic oscillator:

$\left(D^2 + {b \over m} D + \omega_0^2\right) y = 0,$

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:

$\left(D + {b \over 2 m} - \sqrt{{b^2 \over 4 m^2} - \omega_0^2} \right) \left(D + {b \over 2m} + \sqrt{{b^2 \over 4 m^2} - \omega_0^2}\right) y = 0.$

This implies a pair of solutions, one corresponding to

$\left(D + {b \over 2 m} - \sqrt{{b^2 \over 4 m^2} - \omega_0^2} \right) y = 0$

and another to

$\left(D + {b \over 2m} + \sqrt{{b^2 \over 4 m^2} - \omega_0^2}\right) y = 0$

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 / 2m. 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:

$y_H (A_0, A_1) (x) = \left(A_0 \sinh \sqrt{\omega^2 - \omega_0^2} x + A_1 \cosh \sqrt{\omega^2 - \omega_0^2} x\right) e^{-\omega x}.$

However, if |ω| < |ω₀| then it is preferable to get rid of the consequential imaginaries, expressing the general solution as

$y_H (A_0, A_1) (x) = \left(A_0 \sin \sqrt{\omega_0^2 - \omega^2} x + A_1 \cos \sqrt{\omega_0^2 - \omega^2} x\right) e^{-\omega x}.$

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

$\frac {d^{n}y(x)} {dx^{n}} + A_{1}\frac {d^{n-1}y(x)} {dx^{n-1}} + \cdots + A_{n}y(x) = f(x).$

For later convenience, define the characteristic polynomial

$P(v)=v^n+A_1v^{n-1}+\cdots+A_n.$

We find the solution basis $\{y_1(x),y_2(x),\ldots,y_n(x)\}$ as in the homogeneous (f(x)=0) case. We now seek a particular solution y_p(x) by the variation of parameters method. Let the coefficients of the linear combination be functions of x:

$y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x) + \cdots + u_n(x) y_n(x).$

For ease of notation we will drop the dependency on x (i.e. the various (x)). Using the "operator" notation $D = d / d x$ 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

$0=u'_1y_1+u'_2y_2+\cdots+u'_ny_n$

$0=u'_1y'_1+u'_2y'_2+\cdots+u'_ny'_n$

$\cdots$

$0=u'_1y^{(n-2)}_1+u'_2y^{(n-2)}_2+\cdots+u'_ny^{(n-2)}_n$

the parameters commute out, with a little "dirt":

$f=u_1P(D)y_1+u_2P(D)y_2+\cdots+u_nP(D)y_n+u'_1y^{(n-1)}_1+u'_2y^{(n-1)}_2+\cdots+u'_ny^{(n-1)}_n.$

But $P (D) y j = 0$ , therefore

$f=u'_1y^{(n-1)}_1+u'_2y^{(n-1)}_2+\cdots+u'_ny^{(n-1)}_n.$

This, with the constraints, gives a linear system in the $u' j$ . This much can always be solved; in fact, combining Cramer's rule with the Wronskian,

$u'_j=(-1)^{n+j}\frac{W(y_1,\ldots,y_{j-1},y_{j+1}\ldots,y_n)_{0 \choose f}}{W(y_1,y_2,\ldots,y_n)}.$

The rest is a matter of integrating $u' j .$

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 $y'' - 4 y' + 5 y = s i n (k x)$ . We take the solution basis found above ${e (2 + i) x, e (2 - i) x}$ .

$W\,$	$= \begin{vmatrix}e^{(2+i)x}&e^{(2-i)x} \\ (2+i)e^{(2+i)x}&(2-i)e^{(2-i)x} \end{vmatrix}$
	$=e^{4x}\begin{vmatrix}1&1\\ 2+i&2-i\end{vmatrix}$
	$=-2ie^{4x}\,$

$u'_1\,$	$=\frac{1}{W}\begin{vmatrix}0&e^{(2-i)x}\\ \sin(kx)&(2-i)e^{(2-i)x}\end{vmatrix}$
	$=-\frac{i}2\sin(kx)e^{(-2-i)x}$

$u'_2\,$	$=\frac{1}{W}\begin{vmatrix}e^{(2+i)x}&0\\ (2+i)e^{(2+i)x}&\sin(kx)\end{vmatrix}$
	$=\frac{i}{2}\sin(kx)e^{(-2+i)x}.$

Using the list of integrals of exponential functions

$u_1\,$	$=-\frac{i}{2}\int\sin(kx)e^{(-2-i)x}\,dx$
	$=\frac{ie^{(-2-i)x}}{2(3+4i+k^2)}\left((2+i)\sin(kx)+k\cos(kx)\right)$

$u_2\,$	$=\frac i2\int\sin(kx)e^{(-2+i)x}\,dx$
	$=\frac{ie^{(i-2)x}}{2(3-4i+k^2)}\left((i-2)\sin(kx)-k\cos(kx)\right).$

And so

$y_p\,$	$=\frac{i}{2(3+4i+k^2)}\left((2+i)\sin(kx)+k\cos(kx)\right) +\frac{i}{2(3-4i+k^2)}\left((i-2)\sin(kx)-k\cos(kx)\right)$
	$=\frac{(5-k^2)\sin(kx)+4k\cos(kx)}{(3+k^2)^2+16}.$

(Notice that u₁ and u₂ had factors that canceled y₁ and y₂; 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 $c 1 y 1 + c 2 y 2$ is the transient.

Equation with variable coefficients

A linear ODE of order n with variable coefficients has the general form

$p_{n}(x)y^{(n)}(x) + p_{n-1}(x) y^{(n-1)}(x) + \cdots + p_0(x) y(x) = r(x).$

Examples

A simple example is the Cauchy–Euler equation often used in engineering

$x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \cdots + a_0 y(x) = 0.$

First order equation

Examples

Solve the equation

$y'(x)+3y(x)=2 \,$

with the initial condition

$y\left(0\right)=2. \,$

Using the general solution method:

$y=e^{-3x}\left(\int 2 e^{3x}\, dx + \kappa\right). \,$

The indefinite integral is solved to give:

$y=e^{-3x}\left(2/3 e^{3x} + \kappa\right). \,$

Then we can reduce to:

$y=2/3 + \kappa e^{-3x}. \,$

where κ is 4/3 from the initial condition.

A linear ODE of order 1 with variable coefficients has the general form

D y (x) + f (x) y (x) = g (x).

Equations of this form can be solved by multiplying the integrating factor

$e^{\int f(x)\,dx}$

throughout to obtain

$Dy(x)e^{\int f(x)\,dx}+f(x)y(x)e^{\int f(x)\,dx}=g(x)e^{\int f(x) \, dx},$

which simplifies due to the product rule to

$D (y(x)e^{\int f(x)\,dx})=g(x)e^{\int f(x)\,dx}$

which, on integrating both sides, yields

$y(x)e^{\int f(x)\,dx}=\int g(x)e^{\int f(x)\,dx} \,dx+c ~,$

$y(x) = {\int g(x)e^{\int f(x)\,dx} \,dx+c \over e^{\int f(x)\,dx}} ~.$

In other words: The solution of a first-order linear ODE

y'(x) + f (x) y (x) = g (x),

with coefficients that may or may not vary with x, is:

$y=e^{-a(x)}\left(\int g(x) e^{a(x)}\, dx + \kappa\right)$

where $κ$ is the constant of integration, and

$a(x)=\int{f(x)\,dx}.$

Examples

Consider a first order differential equation with constant coefficients:

$\frac{dy}{dx} + b y = 1.$

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

$y(x) = e^{-bx} \left( e^{bx}/b+ C \right) = 1/b + C e^{-bx} .$

Notes

^ Gershenfeld 1999, p.9
^ Robinson 2004, p.5
^ Robinson 2004, p.7

References

Birkhoff, Garret and Rota, Gian-Carlo (1978), Ordinary Differential Equations, New York: John Wiley and Sons, Inc., ISBN 0-471-07411-X
Gershenfeld, Neil (1999), The Nature of Mathematical Modeling, Cambridge, UK.: Cambridge University Press, ISBN 978-0521-570954
Robinson, James C. (2004), An Introduction to Ordinary Differential Equations, Cambridge, UK.: Cambridge University Press, ISBN 0-521-826500

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Linear differential equation

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

Notes

References