In mathematics, a linear differential equation is of the form
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
and, even more precisely by bracketing
The linear operator L may be considered to be of the form^{[1]}
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
where D is the differential operator d/dt (i.e. Dy = y' , D^{2}y = 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
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.
Contents 
The first method of solving linear ordinary differential equations with constant coefficients is due to Euler, who realized that solutions have the form e^{zx}, for possiblycomplex 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
we set y = e^{zx}, leading to
Division by e^{ zx} gives the nthorder polynomial
This equation F(z) = 0, is the "characteristic" equation considered later by Monge and Cauchy.
Formally, the terms
of the original differential equation are replaced by z^{k}. Solving the polynomial gives n values of z, z_{1}, ..., z_{n}. Substitution of any of those values for z into e^{ zx} gives a solution e^{ zix}. 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 

has the characteristic equation This has zeroes, i, −i, and 1 (multiplicity 2). The solution basis is then This corresponds to the realvalued solution basis 
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 , 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 realvalued solutions are generally preferable. Since nonreal roots z then come in conjugate pairs, so do their corresponding basis functions x^{k}e^{ zx}, and the desired result is obtained by replacing each pair with their realvalued 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.
Given . The characteristic equation is which has zeroes 2+i and 2−i. Thus the solution basis {y_{1},y_{2}} is . Now y is a solution if and only if for .
Because the coefficients are real,
The linear combinations
will give us a real basis in {u_{1},u_{2}}.
The second order differential equation
which represents a simple harmonic oscillator, can be restated as
The expression in parenthesis can be factored out, yielding
which has a pair of linearly independent solutions, one for
and another for
The solutions are, respectively,
and
These solutions provide a basis for the twodimensional "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
and
These last two trigonometric solutions are linearly independent, so they can serve as another basis for the solution space, yielding the following general solution:
Given the equation for the damped harmonic oscillator:
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:
Solve using the quadratic formula:
Use these data to factor out the original differential equation:
This implies a pair of solutions, one corresponding to
and another to
The solutions are, respectively,
and
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 twodimensional solution space:
However, if ω < ω_{0} then it is preferable to get rid of the consequential imaginaries, expressing the general solution as
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.
To obtain the solution to the nonhomogeneous 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
We find the solution basis 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:
For ease of notation we will drop the dependency on x (i.e. the various (x)). Using the "operator" notation D = d / dx and a broadminded use of notation, the ODE in question is P(D)y = f; so
With the constraints
the parameters commute out, with a little "dirt":
But P(D)y_{j} = 0, therefore
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,
The rest is a matter of integrating u'_{j}.
The particular solution is not unique; also satisfies the ODE for any set of constants c_{j}.
Suppose y'' − 4y' + 5y = sin(kx). We take the solution basis found above {e^{(2 + i)x},e^{(2 − i)x}}.
Using the list of integrals of exponential functions
And so
(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_{1}y_{1} + c_{2}y_{2} is the transient.
A linear ODE of order n with variable coefficients has the general form
A simple example is the Cauchy–Euler equation often used in engineering
Examples 

Solve
the equation
with the initial condition Using the general solution method: The indefinite integral is solved to give: Then we can reduce to: where κ is 4/3 from the initial condition. 
A linear ODE of order 1 with variable coefficients has the general form
Equations of this form can be solved by multiplying the integrating factor
throughout to obtain
which simplifies due to the product rule to
which, on integrating both sides, yields
In other words: The solution of a firstorder linear ODE
with coefficients that may or may not vary with x, is:
where κ is the constant of integration, and
Consider a first order differential equation with constant coefficients:
This equation is particularly relevant to first order systems such as RC circuits and massdamper systems.
In this case, p(x) = b, r(x) = 1.
Hence its solution is
