Reduction of the Order - Revision history

Khanh at 04:22, 6 November 2021

2021-11-06T04:22:15Z

Khanh at 21:00, 20 October 2021

2021-10-20T21:00:00Z

Lila at 20:50, 20 October 2021

2021-10-20T20:50:23Z

Johnraymond.yanez: Created page with "* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University"

2020-08-12T14:31:21Z

Created page with "* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University"

New page

* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University

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 * [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University
-==References==
+== Licensing ==
-# W. E. Boyce and R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems (8th edition), John Wiley & Sons, Inc., 2005. ISBN 0-471-43338-1.
+Content obtained and/or adapted from:
-# Teschl, Gerald (2012). Ordinary Differential Equations and Dynamical Systems. Providence: American Mathematical Society. ISBN 978-0-8218-8328-0.
+* [https://en.wikipedia.org/wiki/Reduction_of_order Reduction of order, Wikipedia] under a CC BY-SA license

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-'''Reduction of order''' is a technique in [[mathematics]] for solving second-order linear [[Ordinary differential equation|ordinary]] [[differential equation]]s.  It is employed when one solution <math>y_1(x)</math> is known and a second [[Linear independence|linearly independent]] solution <math>y_2(x)</math> is desired. The method also applies to ''n''-th order equations. In this case the [[ansatz]] will yield an (''n''−1)-th order equation for <math>v</math>.
+'''Reduction of order''' is a technique in mathematics for solving second-order linear ordinary differential equations.  It is employed when one solution <math>y_1(x)</math> is known and a second linearly independent solution <math>y_2(x)</math> is desired. The method also applies to ''n''-th order equations. In this case the ansatz will yield an (''n''−1)-th order equation for <math>v</math>.
 == Second-order linear ordinary differential equations==
@@ Line 9: / Line 9: @@
 :<math> a y''(x) + b y'(x) + c y(x) = 0,</math>
-where <math>a, b, c</math> are real non-zero coefficients. Two linearly independent solutions for this ODE can be straightforwardly found using [[characteristic equation (calculus)|characteristic equations]] except for the case when the [[discriminant]], <math>b^2 - 4 a c</math>, vanishes. In this case,
+where <math>a, b, c</math> are real non-zero coefficients. Two linearly independent solutions for this ODE can be straightforwardly found using characteristic equations except for the case when the discriminant, <math>b^2 - 4 a c</math>, vanishes. In this case,
 :<math> a y''(x) + b y'(x) + \frac{b^2}{4a} y(x) = 0,</math>
@@ Line 39: / Line 39: @@
 :<math> a y_1 v'' = 0.</math>
-Since <math>a</math> is assumed non-zero and <math>y_1(x)</math> is an [[exponential function]] (and thus always non-zero), we have
+Since <math>a</math> is assumed non-zero and <math>y_1(x)</math> is an exponential function (and thus always non-zero), we have
 :<math> v'' = 0.</math>
@@ Line 55: / Line 55: @@
 :<math> y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}.</math>
-Finally, we can prove that the second solution <math>y_2(x)</math> found via this method is linearly independent of the first solution by calculating the [[Wronskian]]
+Finally, we can prove that the second solution <math>y_2(x)</math> found via this method is linearly independent of the first solution by calculating the Wronskian
 :<math>W(y_1,y_2)(x) = \begin{vmatrix} y_1 & x y_1 \\ y_1' & y_1 + x y_1' \end{vmatrix} = y_1 ( y_1 + x y_1' ) - x y_1 y_1' = y_1^{2} + x y_1 y_1' - x y_1 y_1' = y_1^{2} = e^{-\frac{b}{a}x} \neq 0.</math>
@@ Line 91: / Line 91: @@
 :<math>v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}.</math>
-The [[integrating factor]] is <math>\mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt}</math>.
+The integrating factor is <math>\mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt}</math>.
 Multiplying the differential equation by the integrating factor <math>\mu(t)</math>, the equation for <math>v(t)</math> can be reduced to
@@ Line 102: / Line 102: @@
 ==Resources==
 * [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University

← Older revision		Revision as of 20:50, 20 October 2021
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		+	'''Reduction of order''' is a technique in [[mathematics]] for solving second-order linear [[Ordinary differential equation\|ordinary]] [[differential equation]]s. It is employed when one solution <math>y_1(x)</math> is known and a second [[Linear independence\|linearly independent]] solution <math>y_2(x)</math> is desired. The method also applies to ''n''-th order equations. In this case the [[ansatz]] will yield an (''n''−1)-th order equation for <math>v</math>.
		+
		+	== Second-order linear ordinary differential equations==
		+
		+	===An example===
		+
		+	Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE)
		+
		+	:<math> a y''(x) + b y'(x) + c y(x) = 0,</math>
		+
		+	where <math>a, b, c</math> are real non-zero coefficients. Two linearly independent solutions for this ODE can be straightforwardly found using [[characteristic equation (calculus)\|characteristic equations]] except for the case when the [[discriminant]], <math>b^2 - 4 a c</math>, vanishes. In this case,
		+
		+	:<math> a y''(x) + b y'(x) + \frac{b^2}{4a} y(x) = 0,</math>
		+
		+	from which only one solution,
		+
		+	:<math>y_1(x) = e^{-\frac{b}{2a} x},</math>
		+
		+	can be found using its characteristic equation.
		+
		+	The method of reduction of order is used to obtain a second linearly independent solution to this differential equation using our one known solution. To find a second solution we take as a guess
		+
		+	:<math>y_2(x) = v(x) y_1(x)</math>
		+
		+	where <math>v(x)</math> is an unknown function to be determined. Since <math>y_2(x)</math> must satisfy the original ODE, we substitute it back in to get
		+
		+	:<math> a \left( v'' y_1 + 2 v' y_1' + v y_1'' \right) + b \left( v' y_1 + v y_1' \right) + \frac{b^2}{4a} v y_1 = 0.</math>
		+
		+	Rearranging this equation in terms of the derivatives of <math>v(x)</math> we get
		+
		+	:<math> \left(a y_1 \right) v'' + \left( 2 a y_1' + b y_1 \right) v' + \left( a y_1'' + b y_1' + \frac{b^2}{4a} y_1 \right) v = 0.</math>
		+
		+	Since we know that <math>y_1(x)</math> is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting <math>y_1(x)</math> into the second term's coefficient yields (for that coefficient)
		+
		+	:<math>2 a \left( - \frac{b}{2a} e^{-\frac{b}{2a} x} \right) + b e^{-\frac{b}{2a} x} = \left( -b + b \right) e^{-\frac{b}{2a} x} = 0.</math>
		+
		+	Therefore, we are left with
		+
		+	:<math> a y_1 v'' = 0.</math>
		+
		+	Since <math>a</math> is assumed non-zero and <math>y_1(x)</math> is an [[exponential function]] (and thus always non-zero), we have
		+
		+	:<math> v'' = 0.</math>
		+
		+	This can be integrated twice to yield
		+
		+	:<math> v(x) = c_1 x + c_2</math>
		+
		+	where <math>c_1, c_2</math> are constants of integration. We now can write our second solution as
		+
		+	:<math> y_2(x) = ( c_1 x + c_2 ) y_1(x) = c_1 x y_1(x) + c_2 y_1(x).</math>
		+
		+	Since the second term in <math>y_2(x)</math> is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of
		+
		+	:<math> y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}.</math>
		+
		+	Finally, we can prove that the second solution <math>y_2(x)</math> found via this method is linearly independent of the first solution by calculating the [[Wronskian]]
		+
		+	:<math>W(y_1,y_2)(x) = \begin{vmatrix} y_1 & x y_1 \\ y_1' & y_1 + x y_1' \end{vmatrix} = y_1 ( y_1 + x y_1' ) - x y_1 y_1' = y_1^{2} + x y_1 y_1' - x y_1 y_1' = y_1^{2} = e^{-\frac{b}{a}x} \neq 0.</math>
		+
		+	Thus <math>y_2(x)</math> is the second linearly independent solution we were looking for.
		+
		+	===General method===
		+
		+	Given the general non-homogeneous linear differential equation
		+
		+	:<math>y'' + p(t)y' + q(t)y = r(t)</math>
		+
		+	and a single solution <math>y_1(t)</math> of the homogeneous equation [<math>r(t)=0</math>], let us try a solution of the full non-homogeneous equation in the form:
		+
		+	:<math>y_2 = v(t)y_1(t)</math>
		+
		+	where <math>v(t)</math> is an arbitrary function. Thus
		+
		+	:<math>y_2' = v'(t)y_1(t) + v(t)y_1'(t)</math>
		+
		+	and
		+
		+	:<math>y_2'' = v''(t)y_1(t) + 2v'(t)y_1'(t) + v(t)y_1''(t).</math>
		+
		+	If these are substituted for <math>y</math>, <math>y'</math>, and <math>y''</math> in the differential equation, then
		+
		+	:<math>y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' + (y_1''(t)+p(t)y_1'(t)+q(t)y_1(t))\,v = r(t).</math>
		+
		+	Since <math>y_1(t)</math> is a solution of the original homogeneous differential equation, <math>y_1''(t)+p(t)y_1'(t)+q(t)y_1(t)=0</math>, so we can reduce to
		+
		+	:<math>y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' = r(t)</math>
		+
		+	which is a first-order differential equation for <math>v'(t)</math> (reduction of order). Divide by <math>y_1(t)</math>, obtaining
		+
		+	:<math>v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}.</math>
		+
		+	The [[integrating factor]] is <math>\mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt}</math>.
		+
		+	Multiplying the differential equation by the integrating factor <math>\mu(t)</math>, the equation for <math>v(t)</math> can be reduced to
		+	:<math>\frac{d}{dt}\left(v'(t) y_1^2(t) e^{\int p(t) dt}\right) = y_1(t)r(t)e^{\int p(t) dt}.</math>
		+
		+	After integrating the last equation, <math>v'(t)</math> is found, containing one constant of integration. Then, integrate <math>v'(t)</math> to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:
		+
		+	:<math>y_2(t) = v(t)y_1(t).</math>
		+
		+
		+	==Resources==
		+	* [https://en.wikipedia.org/wiki/Reduction_of_order Reduction of order], Wikipedia
	* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University		* [https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx Reduction of the Order Notes]. Produced by Paul Dawkins, Lamar University