Linear differential equation

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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 (called the source term). 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 y 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.

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

[edit] Homogeneous equations with constant coefficients

To solve such an equation one makes a substitution

$y = e^{\lambda x}, \!\$

to form the characteristic equation

${\lambda^n +a_{n-1}\lambda^{n-1}+\cdots+a_1\lambda+a_0 = 0}$

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 homogeneous 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

$\frac {d^{n}y} {dx^{n}} + A_{1}\frac {d^{n-1}y} {dx^{n-1}} + \cdots + A_{n}y = 0$

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

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

so dividing by $e z x$ gives the nth-order polynomial

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

In short, the terms

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

of the original differential equation are replaced by z^k. Solving the polynomial gives n values of z, $z_1, \dots,z_n$ . 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.

Example

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

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

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

If z is a (possibly not real) zero(or root) of F(z) of multiplicity m and $k\in\{0,1,\dots,m-1\} \,$ 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 ys; 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.

[edit] 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 iff $y=c_1y_1+c_2y_2 \,$ 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 ${u 1, u 2}$ .

[edit] 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

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

and

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

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).

[edit] 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.

[edit] 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

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

For later convenience, define the characteristic polynomial

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

We find the solution basis $\{y_1,y_2,\ldots,y_n\}$ 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 $D = d / d x$ 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.

[edit] 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.