Talk:Method of variation of parameters

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1 Using this for first-order ODEs
2 u is homogenous solution?
3 Copy from Ordinary differential equation
- 3.1 Method of variation of parameters

[edit] Using this for first-order ODEs

Can't this also be used for 1st order ODE's? Perhaps I'm thinking of a different method, but I was just looking at my Differential Equations Text, and the method seems the same, except for first order equations. Any thoughts? Gershwinrb 06:34, 1 February 2006 (UTC)

Any first-order linear ODE can be solved with little fuss — see Ordinary differential equation#General solution method for first-order linear ODEs — such that techniques like the method of variation of parameters are unnecessary. Ruakh 15:29, 1 February 2006 (UTC)

Should we also have examples for systems of equations and/or higher order ODEs? jleto

[edit] u is homogenous solution?

In the beginning of the Technique section, the article says $u 1$ and $u 2$ are "solutions" to the equation. It really means solutions to the homogenous equation, right? If not, I'm totally confused. This should be changed and made clear. Lavaka 05:32, 15 September 2006 (UTC)

Good call. I've fixed the article now. Ruakh 13:30, 15 September 2006 (UTC)

Thanks! Lavaka 01:55, 20 September 2006 (UTC)

[edit] Copy from Ordinary differential equation

I deleted the following text from the Ordinary differential equation where it consumed far too much space. I copied it here in case anyone is able to salvage some parts and integrate them into this article. MathMartin 20:24, 11 December 2006 (UTC)

[edit] Method of variation of parameters

Main article: Method of variation of parameters

As explained above, the general solution to a non-homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ can be expressed as the sum of the general solution $y h (x)$ to the corresponding homogenous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ and any one solution $y p (x)$ to $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ .

Like the method of undetermined coefficients, described above, the method of variation of parameters is a method for finding one solution to $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ , having already found the general solution to $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ . Unlike the method of undetermined coefficients, which fails except with certain specific forms of g(x), the method of variation of parameters will always work; however, it is significantly more difficult to use.

For a second-order equation, the method of variation of parameters makes use of the following fact:

[edit] Fact

Let p(x), q(x), and g(x) be functions, and let $y 1 (x)$ and $y 2 (x)$ be solutions to the homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ . Further, let u(x) and v(x) be functions such that $u'(x) y 1 (x) + v'(x) y 2 (x) = 0$ and $u'(x) y 1'(x) + v'(x) y 2'(x) = g (x)$ for all x, and define $y p (x) = u (x) y 1 (x) + v (x) y 2 (x)$ . Then $y p (x)$ is a solution to the non-homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ .

[edit] Proof

$y p (x) = u (x) y 1 (x) + v (x) y 2 (x)$

$y p'(x)$	$= u'(x) y 1 (x) + u (x) y 1'(x) + v'(x) y 2 (x) + v (x) y 2'(x)$
	$= 0 + u (x) y 1'(x) + v (x) y 2'(x)$

$y p''(x)$	$= u'(x) y 1'(x) + u (x) y 1''(x) + v'(x) y 2'(x) + v (x) y 2''(x)$
	$= g (x) + u (x) y 1''(x) + v (x) y 2''(x)$

$y p''(x) + p (x) y' p (x) + q (x) y p (x) = g (x) + u (x) y 1''(x) + v (x) y 2''(x) + p (x) u (x) y 1'(x) + p (x) v (x) y 2'(x) + q (x) u (x) y 1 (x) + q (x) v (x) y 2 (x)$

$= g (x) + u (x)(y 1''(x) + p (x) y 1'(x) + q (x) y 1 (x)) + v (x)(y 2''(x) + p (x) y 2'(x) + q (x) y 2 (x)) = g (x) + 0 + 0 = g (x)$

[edit] Usage

To solve the second-order, non-homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ using the method of variation of parameters, use the following steps:

Find the general solution to the corresponding homogeneous equation $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ . Specifically, find two linearly independent solutions $y 1 (x)$ and $y 2 (x)$ .
Since $y 1 (x)$ and $y 2 (x)$ are linearly independent solutions, their Wronskian $y 1 (x) y 2'(x) - y 1'(x) y 2 (x)$ is nonzero, so we can compute $- (g (x) y 2 (x)) / (y 1 (x) y 2'(x) - y 1'(x) y 2 (x))$ and $(g (x) y 1 (x)) / (y 1 (x) y 2'(x) - y 1'(x) y 2 (x))$ . If the former is equal to u'(x) and the latter to v'(x), then u and v satisfy the two constraints given above: that $u'(x) y 1 (x) + v'(x) y 2 (x) = 0$ and that $u'(x) y 1'(x) + v'(x) y 2'(x) = g (x)$ . We can tell this after multiplying by the denominator and comparing coefficients.
Integrate $- (g (x) y 2 (x)) / (y 1 (x) y 2'(x) - y 1'(x) y 2 (x))$ and $(g (x) y 1 (x)) / (y 1 (x) y 2'(x) - y 1'(x) y 2 (x))$ to obtain u(x) and v(x), respectively. (Note that we only need one choice of u and v, so there is no need for constants of integration.)
Compute $y p (x) = u (x) y 1 (x) + v (x) y 2 (x)$ . The function $y p$ is one solution of $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ .
The general solution is $c 1 y 1 (x) + c 2 y 2 (x) + y p (x)$ , where $c 1$ and $c 2$ are arbitrary constants.

[edit] Higher-order equations

The method of variation of parameters can also be used with higher-order equations. For example, if $y 1 (x)$ , $y 2 (x)$ , and $y 3 (x)$ are linearly independent solutions to $y'''(x) + p (x) y''(x) + q (x) y'(x) + r (x) y (x) = 0$ , then there exist functions u(x), v(x), and w(x) such that $u'(x) y 1 (x) + v'(x) y 2 (x) + w'(x) y 3 (x) = 0$ , $u'(x) y 1'(x) + v'(x) y 2'(x) + w'(x) y 3'(x) = 0$ , and $u'(x) y 1''(x) + v'(x) y 2''(x) + w'(x) y 3''(x) = g (x)$ . Having found such functions (by solving algebraically for u'(x), v'(x), and w'(x), then integrating each), we have $y p (x) = u (x) y 1 (x) + v (x) y 2 (x) + w (x) y 3 (x)$ , one solution to the equation $y'''(x) + p (x) y''(x) + q (x) y'(x) + r (x) y (x) = g (x)$ .

[edit] Example

Solve the previous example, $y'' + y = sec x$ Recall $\sec x = \frac{1}{{\cos x}} = f$ . From technique learned from 3.1, LHS has root of $r = \pm i$ that yield $y c = C 1 cos x + C 2 sin x$ , (so $y 1 = cos x$ , $y 2 = sin x$ ) and its derivatives

$\left\{ {\begin{matrix} {\dot u = \frac{{ - y_2 f}}{W} = \frac{{ - \sin x}}{{\cos x}} = \tan x} \\ {\dot v = \frac{{y_1 f}}{W} = \frac{{\cos x}}{{\cos x}} = 1} \\ \end{matrix}} \right.$

where the Wronskian

$W\left( {y_1,y_2 :x} \right) = \left| {\begin{matrix} {\cos x} & {\sin x} \\ { - \sin x} & {\cos x} \\ \end{matrix}} \right| = 1$

were computed in order to seek solution to its derivatives.

Upon integration,

$\left\{ \begin{matrix} u = - \int {\tan x\,dx = - \ln \left| {\sec x} \right| + C} \\ v = \int {1\,dx = x + C} \\ \end{matrix} \right.$

Computing $y p$ and $y G$ :

$\begin{matrix} y_p = f = uy_1 + vy_2 = \cos x\ln \left| {\cos x} \right| + x\sin x \\ y_G = y_c + y_p = C_1 \cos x + C_2 \sin x + x\sin x + \cos x\ln \left( {\cos x} \right) \\ \end{matrix}$