Linear Differential Equations

In mathematics, a linear differential equation is a differential equation that is defined by a linear polynomial in the unknown function and its derivatives, that is an equation of the form

$a_{0}(x)y+a_{1}(x)y'+a_{2}(x)y''\cdots +a_{n}(x)y^{(n)}=b(x)$

where a₀(x), …, a_n(x) and b(x) are arbitrary differentiable functions that do not need to be linear, and y′, …, y⁽ⁿ⁾ are the successive derivatives of an unknown function y of the variable x.

Such an equation is an ordinary differential equation (ODE). A linear differential equation may also be a linear partial differential equation (PDE), if the unknown function depends on several variables, and the derivatives that appear in the equation are partial derivatives.

A linear differential equation or a system of linear equations such that the associated homogeneous equations have constant coefficients may be solved by quadrature, which means that the solutions may be expressed in terms of integrals. This is also true for a linear equation of order one, with non-constant coefficients. An equation of order two or higher with non-constant coefficients cannot, in general, be solved by quadrature. For order two, Kovacic's algorithm allows deciding whether there are solutions in terms of integrals, and computing them if any.

The solutions of linear differential equations with polynomial coefficients are called holonomic functions. This class of functions is stable under sums, products, differentiation, integration, and contains many usual functions and special functions such as exponential function, logarithm, sine, cosine, inverse trigonometric functions, error function, Bessel functions and hypergeometric functions. Their representation by the defining differential equation and initial conditions allows making algorithmic (on these functions) most operations of calculus, such as computation of antiderivatives, limits, asymptotic expansion, and numerical evaluation to any precision, with a certified error bound.

Basic terminology

The highest order of derivation that appears in a (linear) differential equation is the order of the equation. The term b(x), which does not depend on the unknown function and its derivatives, is sometimes called the constant term of the equation (by analogy with algebraic equations), even when this term is a non-constant function. If the constant term is the zero function, then the differential equation is said to be homogeneous, as it is a homogeneous polynomial in the unknown function and its derivatives. The equation obtained by replacing, in a linear differential equation, the constant term by the zero function is the associated homogeneous equation. A differential equation has constant coefficients if only constant functions appear as coefficients in the associated homogeneous equation.

A solution of a differential equation is a function that satisfies the equation. The solutions of a homogeneous linear differential equation form a vector space. In the ordinary case, this vector space has a finite dimension, equal to the order of the equation. All solutions of a linear differential equation are found by adding to a particular solution any solution of the associated homogeneous equation.

Linear differential operator

A basic differential operator of order i is a mapping that maps any differentiable function to its ith derivative, or, in the case of several variables, to one of its partial derivatives of order i. It is commonly denoted

${\frac {d^{i}}{dx^{i}}}$

in the case of univariate functions, and

${\frac {\partial ^{i_{1}+\cdots +i_{n}}}{\partial x_{1}^{i_{1}}\cdots \partial x_{n}^{i_{n}}}}$

in the case of functions of n variables. The basic differential operators include the derivative of order 0, which is the identity mapping.

A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. In the univariate case, a linear operator has thus the form

$a_{0}(x)+a_{1}(x){\frac {d}{dx}}+\cdots +a_{n}(x){\frac {d^{n}}{dx^{n}}},$

where a₀(x), …, a_n(x) are differentiable functions, and the nonnegative integer n is the order of the operator (if a_n(x) is not the zero function).

Let L be a linear differential operator. The application of L to a function f is usually denoted Lf or Lf(X), if one needs to specify the variable (this must not be confused with a multiplication). A linear differential operator is a linear operator, since it maps sums to sums and the product by a scalar to the product by the same scalar.

As the sum of two linear operators is a linear operator, as well as the product (on the left) of a linear operator by a differentiable function, the linear differential operators form a vector space over the real numbers or the complex numbers (depending on the nature of the functions that are considered). They form also a free module over the ring of differentiable functions.

The language of operators allows a compact writing for differentiable equations: if

$L=a_{0}(x)+a_{1}(x){\frac {d}{dx}}+\cdots +a_{n}(x){\frac {d^{n}}{dx^{n}}},$

is a linear differential operator, then the equation

$a_{0}(x)y+a_{1}(x)y'+a_{2}(x)y''+\cdots +a_{n}(x)y^{(n)}=b(x)$

may be rewritten

$Ly=b(x).$

There may be several variants to this notation; in particular the variable of differentiation may appear explicitly or not in y and the right-hand and of the equation, such as Ly(x) = b(x) or Ly = b.

The kernel of a linear differential operator is its kernel as a linear mapping, that is the vector space of the solutions of the (homogeneous) differential equation Ly = 0.

In the case of an ordinary differential operator of order n, Carathéodory's existence theorem implies that, under very mild conditions, the kernel of L is a vector space of dimension n, and that the solutions of the equation Ly(x) = b(x) have the form

$S_{0}(x)+c_{1}S_{1}(x)+\cdots +c_{n}S_{n}(x),$

where c₁, …, c_n are arbitrary numbers. Typically, the hypotheses of Carathéodory's theorem are satisfied in an interval I, if the functions b, a₀, …, a_n are continuous in I, and there is a positive real number k such that |a_n(x)| > k for every x in I.

Homogeneous equation with constant coefficients

A homogeneous linear differential equation has constant coefficients if it has the form

$a_{0}y+a_{1}y'+a_{2}y''+\cdots +a_{n}y^{(n)}=0$

where a₁, …, a_n 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 e^x, which is the unique solution of the equation f′ = f such that f(0) = 1. It follows that the nth derivative of e^cx is cⁿe^cx, and this allows solving homogeneous linear differential equations rather easily.

Let

$a_{0}y+a_{1}y'+a_{2}y''+\cdots +a_{n}y^{(n)}=0$

be a homogeneous linear differential equation with constant coefficients (that is a₀, …, a_n are real or complex numbers).

Searching solutions of this equation that have the form e^αx is equivalent to searching the constants α such that

$a_{0}e^{\alpha x}+a_{1}\alpha e^{\alpha x}+a_{2}\alpha ^{2}e^{\alpha x}+\cdots +a_{n}\alpha ^{n}e^{\alpha x}=0.$

Factoring out e^αx (which is never zero), shows that α must be a root of the characteristic polynomial

$a_{0}+a_{1}t+a_{2}t^{2}+\cdots +a_{n}t^{n}$

of the differential equation, which is the left-hand side of the characteristic equation

$a_{0}+a_{1}t+a_{2}t^{2}+\cdots +a_{n}t^{n}=0.$

When these roots are all distinct, one has 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 x = 0, …, n – 1. Together they form a basis of the vector space of solutions of the differential equation (that is, the kernel of the differential operator).

Example
$y''''-2y'''+2y''-2y'+y=0$ has the characteristic equation $z^{4}-2z^{3}+2z^{2}-2z+1=0.$ This has zeros, i, −i, and 1 (multiplicity 2). The solution basis is thus $e^{ix},\;e^{-ix},\;e^{x},\;xe^{x}.$ A real basis of solution is thus $\cos x,\;\sin x,\;e^{x},\;xe^{x}.$

In the case where the characteristic polynomial has only simple roots, the preceding provides a complete basis of the solutions vector space. In the case of multiple roots, more linearly independent solutions are needed for having a basis. These have the form

$x^{k}e^{\alpha x},$

where k is a nonnegative integer, α is a root of the characteristic polynomial of multiplicity m, and k < m. For proving that these functions are solutions, one may remark that if α is a root of the characteristic polynomial of multiplicity m, the characteristic polynomial may be factored as P(t)&hairsp;(t − α)^m. Thus, applying the differential operator of the equation is equivalent with applying first m times the operator ${\textstyle {\frac {d}{dx}}-\alpha }$, and then the operator that has P as characteristic polynomial. By the exponential shift theorem,

$\left({\frac {d}{dx}}-\alpha \right)\left(x^{k}e^{\alpha x}\right)=kx^{k-1}e^{\alpha x},$

and thus one gets zero after k + 1 application of ${\textstyle {\frac {d}{dx}}-\alpha }$.

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 base 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 functions. Such a basis may be obtained from the preceding basis by remarking that, if a + ib is a root of the characteristic polynomial, then a – ib is also a root, of the same multiplicity. Thus a real basis is obtained by using Euler's formula, and replacing x^ke^(a+ib)x and x^ke^(a−ib)x by x^ke^ax cos(bx) and x^ke^ax sin(bx).

Second-order case

A homogeneous linear differential equation of the second order may be written

$y''+ay'+by=0,$

and its characteristic polynomial is

$r^{2}+ar+b.$

If a and b are real, there are three cases for the solutions, depending on the discriminant D = a² − 4b. In all three cases, the general solution depends on two arbitrary constants c₁ and c₂.

• If D > 0, the characteristic polynomial has two distinct real roots α, and β. In this case, the general solution is

$c_{1}e^{\alpha x}+c_{2}e^{\beta x}.$

• If D = 0, the characteristic polynomial has a double root −a/2, and the general solution is

$(c_{1}+c_{2}x)e^{-ax/2}.$

• If D < 0, the characteristic polynomial has two complex conjugate roots α ± βi, and the general solution is

$c_{1}e^{(\alpha +\beta i)x}+c_{2}e^{(\alpha -\beta i)x},$

which may be rewritten in real terms, using Euler's formula as

$e^{\alpha x}(c_{1}\cos(\beta x)+c_{2}\sin(\beta x)).$

Finding the solution y(x) satisfying y(0) = d₁ and y′(0) = d₂, one equates the values of the above general solution at 0 and its derivative there to d₁ and d₂, respectively. This results in a linear system of two linear equations in the two unknowns c₁ and c₂. Solving this system gives the solution for a so-called Cauchy problem, in which the values at 0 for the solution of the DEQ and its derivative are specified.

Non-homogeneous equation with constant coefficients

A non-homogeneous equation of order n with constant coefficients may be written

$y^{(n)}(x)+a_{1}y^{(n-1)}(x)+\cdots +a_{n-1}y'(x)+a_{n}y(x)=f(x),$

where a₁, …, a_n are real or complex numbers, f is a given function of x, and y is the unknown function (for sake of simplicity, "(x)" will be omitted in the following).

There are several methods for solving such an equation. The best method depends on the nature of the function f that makes the equation non-homogeneous. If f is a linear combination of exponential and sinusoidal functions, then the exponential response formula may be used. If, more generally, f is a linear combination of functions of the form xⁿe^ax, xⁿ cos(ax), and xⁿ sin(ax), where n is a nonnegative integer, and a a constant (which need not be the same in each term), then the method of undetermined coefficients may be used. Still more general, the annihilator method applies when f satisfies a homogeneous linear differential equation, typically, a holonomic function.

The most general method is the variation of constants, which is presented here.

The general solution of the associated homogeneous equation

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

is

$y=u_{1}y_{1}+\cdots +u_{n}y_{n},$

where (y₁, …, y_n) is a basis of the vector space of the solutions and u₁, …, u_n are arbitrary constants. The method of variation of constants takes its name from the following idea. Instead of considering u₁, …, u_n as constants, they can considered as unknown functions that have to be determined for making y a solution of the non-homogeneous equation. For this purpose, one adds the constraints

${\begin{aligned}0&=u'_{1}y_{1}+u'_{2}y_{2}+\cdots +u'_{n}y_{n}\\0&=u'_{1}y'_{1}+u'_{2}y'_{2}+\cdots +u'_{n}y'_{n}\\&\;\;\vdots \\0&=u'_{1}y_{1}^{(n-2)}+u'_{2}y_{2}^{(n-2)}+\cdots +u'_{n}y_{n}^{(n-2)},\end{aligned}}$

which imply (by product rule and induction)

$y^{(i)}=u_{1}y_{1}^{(i)}+\cdots +u_{n}y_{n}^{(i)}$

for i = 1, …, n – 1, and

$y^{(n)}=u_{1}y_{1}^{(n)}+\cdots +u_{n}y_{n}^{(n)}+u'_{1}y_{1}^{(n-1)}+u'_{2}y_{2}^{(n-1)}+\cdots +u'_{n}y_{n}^{(n-1)}.$

Replacing in the original equation y and its derivatives by these expressions, and using the fact that y₁, …, y_n are solutions of the original homogeneous equation, one gets

$f=u'_{1}y_{1}^{(n-1)}+\cdots +u'_{n}y_{n}^{(n-1)}.$

This equation and the above ones with 0 as left-hand side form a system of n linear equations in u′₁, …, u′_n whose coefficients are known functions (f, the $y_{i}$, and their derivatives). This system can be solved by any method of linear algebra. The computation of antiderivatives gives u₁, …, u_n, and then y = u₁y₁ + ⋯ + u_ny_n.

As antiderivatives are defined up to the addition of a constant, one finds again that the general solution of the non-homogeneous equation is the sum of an arbitrary solution and the general solution of the associated homogeneous equation.

First-order linear equation

The general form of a linear ordinary differential equation of order 1, after dividing out the coefficient of y′(x), is:

$y'(x)=f(x)y(x)+g(x).$

If the equation is homogeneous, i.e. g(x) = 0, one may rewrite and integrate:

${\frac {y'}{y}}=f,\qquad \log y=k+F,$

where k is an arbitrary constant of integration and $F=\textstyle \int f\,dx$ is any antiderivative of f. Thus, the general solution of the homogeneous equation is

$y=ce^{F},$

where c = e^k is an arbitrary constant.

For the general non-homogeneous equation, one may multiply it by the reciprocal e^−F of a solution of the homogeneous equation. This gives

$y'e^{-F}-yfe^{-F}=ge^{-F}.$

As $-fe^{-F}={\frac {d}{dx}}\left(e^{-F}\right),$ the product rule allows rewriting the equation as

${\frac {d}{dx}}\left(ye^{-F}\right)=ge^{-F}.$

Thus, the general solution is

$y=ce^{F}+e^{F}\int ge^{-F}dx,$

where c is a constant of integration, and F is any antiderivative of f (changing of antiderivative amounts to change the constant of integration).

Example

Solving the equation

$y'(x)+{\frac {y(x)}{x}}=3x.$

The associated homogeneous equation $y'(x)+{\frac {y(x)}{x}}=0$ gives

${\frac {y'}{y}}=-{\frac {1}{x}},$

that is

$y={\frac {c}{x}}.$

Dividing the original equation by one of these solutions gives

$xy'+y=3x^{2}.$

That is

$(xy)'=3x^{2},$

$xy=x^{3}+c,$

and

$y(x)=x^{2}+c/x.$

For the initial condition

$y(1)=\alpha ,$

one gets the particular solution

$y(x)=x^{2}+{\frac {\alpha -1}{x}}.$

Making Linear Equations from Non-Linear Equations

Sometimes a non-linear equation, which is not solvable like this, can be made linear, and more easily solvable, by applying a substitution.

Example

$y{\frac {dy}{dx}}+xy^{2}=5x$

Let's make the following substitution:

$v=y^{2}\,$

${\frac {dv}{dx}}=2y{\frac {dy}{dx}}$

Plugging in, we get

${\frac {1}{2}}{\frac {dv}{dx}}+xv=5x$

${\frac {dv}{dx}}+2xv=10x$

We can then solve as a linear equation in v, using the step-by-step method above:

Step 1: Find the integrating factor:

$e^{\int P(x)dx}$

$e^{\int 2xdx}=e^{x^{2}+C}$

$e^{\int P(x)dx}=Ce^{x^{2}}$

Letting C=1 for convenience, we get $e^{x^{2}}$ as our integrating factor.

Step 2: Multiply through

$e^{x^{2}}{\frac {dv}{dx}}+e^{x^{2}}xv=10e^{x^{2}}x$

Step 3: Recognize that the left hand is ${\frac {d}{dx}}e^{\int P(x)dx}v$

${\frac {d}{dx}}e^{x^{2}}v=10e^{x^{2}}x$

Step 4: Integrate both sides w.r.t.x.

$\int ({\frac {d}{dx}}e^{x^{2}}v)dx=\int 10e^{x^{2}}xdx$

$e^{x^{2}}v=5e^{x^{2}}+C$

Step 5: Solve for v.

$v=5+{\frac {C}{e^{x^{2}}}}$

Now that we have v, solve for y.

$v=y^{2}\,$

$y^{2}=5+{\frac {C}{e^{x^{2}}}}\,$

Higher order linear equations

A linear ordinary equation of order one with variable coefficients may be solved by quadrature, which means that the solutions may be expressed in terms of integrals. This is not the case for order at least two. This is the main result of Picard–Vessiot theory which was initiated by Émile Picard and Ernest Vessiot, and whose recent developments are called differential Galois theory.

The impossibility of solving by quadrature can be compared with the Abel–Ruffini theorem, which states that an algebraic equation of degree at least five cannot, in general, be solved by radicals. This analogy extends to the proof methods and motivates the denomination of differential Galois theory.

Similarly to the algebraic case, the theory allows deciding which equations may be solved by quadrature, and if possible solving them. However, for both theories, the necessary computations are extremely difficult, even with the most powerful computers.

Nevertheless, the case of order two with rational coefficients has been completely solved by Kovacic's algorithm.

Cauchy–Euler equation

Cauchy–Euler equations are examples of equations of any order, with variable coefficients, that can be solved explicitly. These are the equations of the form

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

where $a_{0},\ldots ,a_{n-1}$ are constant coefficients.

Licensing

Content obtained and/or adapted from:

• Linear differential equation, Wikipedia under a CC BY-SA license
• First Order Linear 1, Wikibooks: Differential Equations under a CC BY-SA license