What is a linear ordinary differential equation

Click For Summary
SUMMARY

A linear ordinary differential equation (ODE) is defined as an equation of the form ∑_{i=0}^n a_i(x)y^{(i)}(x) = b(x), where y^{(i)}(x) represents the ith derivative of y. The general solution consists of a particular solution and a linear combination of n independent solutions from the homogeneous part, ∑_{i=0}^n a_i(x)y^{(i)}(x) = 0. The method of undetermined coefficients is used to find particular solutions, while the characteristic polynomial method is employed for homogeneous solutions. An example demonstrates the general solution for the equation y'' - 2y' - 8y = \sin{x} + \cos{x}.

PREREQUISITES
  • Understanding of linear ordinary differential equations
  • Familiarity with the method of undetermined coefficients
  • Knowledge of characteristic polynomials
  • Basic calculus, including differentiation and integration
NEXT STEPS
  • Study the method of characteristic polynomials in detail
  • Explore the method of undetermined coefficients for various types of functions
  • Learn about complex roots in differential equations
  • Investigate repeated roots and their impact on general solutions
USEFUL FOR

Mathematicians, engineering students, and anyone involved in solving differential equations will benefit from this discussion, particularly those focusing on linear ODEs and their applications.

Messages
19,910
Reaction score
10,917
Definition/Summary

An nth-order linear ordinary differential equation (ODE) is a differential equation of the form
\sum_{i=0}^n a_i(x)y^{(i)}(x)\ =\ b(x)
where y^{(i)}(x) denotes the ith derivative of y with respect to x.

The difference between any two solutions is a solution of the homogeneous part:
\sum_{i=0}^n a_i(x)y^{(i)}(x)\ =\ 0

This equation has n independent solutions, and every solution is a linear combination of them.

So the general solution of a linear ODE is the sum of one particular solution of the whole equation plus a linear combination of n independent solutions of the homogeneous part (in other words: a particular solution plus any homogeneous solution).

Equations
Extended explanation


Homogeneous linear ODE:

A homogeneous linear ODE is a linear ODE in which b(x)=0.

Homogeneous solutions:

These are linear combinations of solutions of the form e^{\lambda x} or x^ie^{\lambda x}.

Any solution of the form e^{\lambda x} can be found by the method shown in the following example.

For the general method of solution, see the method of characteristic polynomial, at the foot of this page.

Example of homogeneous solution:

Find the general solution to y'' - 2y' -8y = \sin{x}+ \cos{x}

First we must find the general homogeneous solution, y_h.

Let's guess that y_h is in the form y_h=e^{\lambda x} - this seems like
a good guess, doesn't it? So, plugging in, we have
\left(\lambda^2 - 2\lambda -8\right) e^{\lambda x} = 0
Now, since e^{\lambda x} is never 0, we can solve the above equation as a quadratic (the characteristic quadratic),
and we find that either \lambda=4 or \lambda=-2.
Therefore the general solution is
y_h = C_1 e^{4x} + C_2 e^{-2x}
where C_1 and C_2 are arbitrary constants.

Particular solutions:

We only need one particular solution.

Finding one is generally a combination of common-sense and guesswork, based on the nature of the polynomial b(x).

Let's see how to do it with the above example, in which b(x)\ =\ \sin{x}+ \cos{x}, using the method of undetermined coefficients …

There is also the more general variation of parameters method, but this is not usually needed in examination questions.

Example of Particular Solution by the Method of Undetermined Coefficients:

Let's guess a particular solution in the form y_p\ =\ A\sin{x} + B\cos{x}.

The method of undetermined coefficients is to solve for A and B by plugging y\ =\ y_p into the original equation, giving:

\sin{x}(-9A\ +\ 2B)\ +\ \cos{x}(-2A\ -\ 9B)\ =\ \sin{x}\ +\ \cos{x}

which, dealing with the coefficients of \sin{x} and \cos{x} separately, gives:

A\ =\ -11/85\ \ \ \ B = -7/85

and so y_p\ =\ -(11\sin{x}\ +\ 7\cos{x})/85 is a solution to the original equation.

Using this as the particular solution, the general solution of the original equation is:

y= C_1 e^{4x} + C_2 e^{-2x} - (11\sin{x}\ +\ 7\cos{x})/85

Homogeneous solution by characteristic polynomial:

This is not the same as the characteristic polynomial of a matrix or matroid

In a linear differential equation, the derivative may be replaced by an operator, D, giving a polynomial equation in D:

\sum_{n\,=\,0}^m\,a_n\,\frac{d^ny}{dx^n}\ =\ 0\ \mapsto\ \left(\sum_{n\,=\,0}^m\,a_n\,D^n\right)y\ =\ 0

If this polynomial has distinct (different) roots \lambda_1,\dots,\lambda_m:

\prod_{n\,=\,1}^m(D\,-\,\lambda_n)\ =\ 0

then the general solution is a linear combination of the solutions of each of the equations:

\left(D\,-\,\lambda_n\right)y\ =\ 0

which are the same as \frac{dy}{dx}\ =\ \lambda_n\,y

and so the general solution is of the form:

y\ =\ \sum_{n\,=\,1}^m\,C_n\,e^{\lambda_nx}

For a pair of complex roots (they always come in conjugate pairs) p\ \pm\ iq or r\,e^{\pm is}, a pair of C_ne^{r_nx} may be replaced by e^{px}(A\,cos(qx)\,+\,iB\,sin(qx)) or r^k(A\,cos(sk)\,+\,iB\,sin(sk))

However, if the polynomial has some repeated roots:

\prod_{p\,=\,1}^q\prod_{n\,=\,1}^p(D\,-\,\lambda_n)^p\,y\ =\ 0

then the general solution is of the form:

y\ =\ \sum_{p\,=\,1}^q\sum_{n\,=\,1}^pC_{n,p}\,x^{p-1}\,e^{\lambda_nx}

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
Last edited by a moderator:
Physics news on Phys.org

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
867
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 52 ·
2
Replies
52
Views
9K
  • · Replies 3 ·
Replies
3
Views
3K