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An nth-order linear ordinary differential equation (ODE) is a differential equation of the form

[tex]\sum_{i=0}^n a_i(x)y^{(i)}(x)\ =\ b(x)[/tex]

where [itex]y^{(i)}(x)[/itex] denotes the ith derivative of y with respect to x.

The difference between any two solutions is a solution of the homogeneous part:

[tex]\sum_{i=0}^n a_i(x)y^{(i)}(x)\ =\ 0[/tex]

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 [itex]b(x)=0.[/itex]

Homogeneous solutions:

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

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

For thegeneralmethod of solution, see the method ofcharacteristic polynomial, at the foot of this page.

Example of homogeneous solution:

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

First we must find the general homogeneous solution, [itex]y_h[/itex].

Let's guess that [itex]y_h[/itex] is in the form [itex]y_h=e^{\lambda x}[/itex] - this seems like

a good guess, doesn't it? So, plugging in, we have

[tex]\left(\lambda^2 - 2\lambda -8\right) e^{\lambda x} = 0[/tex]

Now, since [itex]e^{\lambda x}[/itex] is never [itex]0[/itex], we can solve the above equation as a quadratic (the characteristic quadratic),

and we find that either [itex]\lambda=4[/itex] or [itex]\lambda=-2[/itex].

Therefore the general solution is

[tex]y_h = C_1 e^{4x} + C_2 e^{-2x}[/tex]

where [itex]C_1[/itex] and [itex]C_2[/itex] are arbitrary constants.

Particular solutions:

We only needoneparticular solution.

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

Let's see how to do it with the above example, in which [itex]b(x)\ =\ \sin{x}+ \cos{x}[/itex], 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 [itex]y_p\ =\ A\sin{x} + B\cos{x}[/itex].

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

[tex]\sin{x}(-9A\ +\ 2B)\ +\ \cos{x}(-2A\ -\ 9B)\ =\ \sin{x}\ +\ \cos{x}[/tex]

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

[tex]A\ =\ -11/85\ \ \ \ B = -7/85[/tex]

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

Using this as theparticular solution, thegeneral solutionof the original equation is:

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

Homogeneous solution by characteristic polynomial:

This isnotthe 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:

[tex]\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[/tex]

If this polynomial hasdistinct(different) roots [itex]\lambda_1,\dots,\lambda_m[/itex]:

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

then thegeneral solutionis a linear combination of the solutions of each of the equations:

[tex]\left(D\,-\,\lambda_n\right)y\ =\ 0[/tex]

which are the same as [tex]\frac{dy}{dx}\ =\ \lambda_n\,y[/tex]

and so the general solution is of the form:

[tex]y\ =\ \sum_{n\,=\,1}^m\,C_n\,e^{\lambda_nx}[/tex]

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

However, if the polynomial has somerepeatedroots:

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

then the general solution is of the form:

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

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# What is a linear ordinary differential equation

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