Linear differential equations are those that directly relates a function to how the function changes by just a number (although that number can change with time). Many examples exist in the real world: exponential growth, Newton's Law of Cooling, electrical engineering, etc. Linear differential equations are the easiest type of differential equations to solve, too, so many more difficult problems are frequently linearized, allowing them to be approximated by the solution to a linear DE. Linear ODEs are easier to deal with than linear PDEs, so I'll only be talking about ODEs.

An nth-order linear differential equation is one of the form L[y] = y(n) + f1y(n-1) + ... + fn-1y' + fny = g, where fi, g and y are functions of t. It is linear because L is a linear operator (i.e. if α and β are constants and y and z are functions of t, L[αy + βz]=αL[y] + βL[z]). Explicit solutions of these equations cannot necessarily be computed, and the likelihood of finding one becomes lower as the order gets higher. Most of the time, though, an explicit solution can be found for first-order linear ODEs. For a detailed explanation, see the writeups in differential equation, but the general solution of y' + fy = g is y = exp{-∫f⋅dt}⋅∫exp{∫f⋅dt}⋅g dt. Since these integrals cannot always be evaluated, explicit solutions do not always exist.

The simplest case is the constant coefficient equation, for which a solution can be found for any order of ODE. Let L[y] = y(n) + f1y(n-1) + ... + fn-1y' + fny = g. When the fi are all constants, then the solution to the homogeneous (i.e. g=0) equation is a linear combination of solutions yi (i=1,...,n), where the yi are exponentials and the exponents are the roots of the characteristic polynomial of the equation. If a root r is repeated with multiplicity k, then the solutions are of the form:

y1 = ert,
y2 = t⋅ert,
y3 = t2⋅ert,
yk = tk-1⋅ert


y''' - 4y'' + 5y' - 2 = 0

r3 - 4r2 + 5r - 2 = 0   is the characteristic equation. Solving...

r1=1, r2=1, r3=2   are the roots of the polynomial. Note r1=r2=1.

y1 = e1⋅t, y2 = t⋅e1⋅t, y3 = e2⋅t

The solution is y = c1y1 + c2y2 + c3y3 = c1e1⋅t + c2t⋅e1⋅t + c3e2⋅t

After solving the homogeneous problem in all of these cases, the particular solution can be found to the nonhomogeneous (g ≠ 0) problem via variation of parameters. Finding solutions to second-order linear ODEs can be very difficult and often impossible. Some techniques for solving these equations include converting it to a first-order exact linear ODE, reduction of order (if one solution can be found), using a Laplace Transfrom, and series solutions. As difficult as it is to solve these linear problems, it is much, much harder to solve nonlinear problems. For problems for which solutions can't be found, people usually use numerical methods (like the Runge-Kutta methods) to approximate the solution .

Log in or register to write something here or to contact authors.