Reduction of order is a technique in mathematics for solving second-order 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.
A Simple Example
Consider the general second-order constant coefficient ODE
- <math> a y(x) + b y'(x) + c y(x) = 0, \;</math>
where <math>a, b, c</math> are real non-zero coefficients. Furthermore, assume that the associated characteristic equation
- <math> a \lambda^{2} + b \lambda + c = 0 \;</math>
has repeated roots (i.e. the discriminant, <math>b^2 - 4 a c</math>, vanishes). Thus we have
- <math> \lambda_{1,2} = -\frac{b}{2 a}.</math>
Thus our one solution to the ODE is
- <math>y_1(x) = e^{-\frac{b}{2 a} x}.</math>
To find a second solution we take as an ansatz
- <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) + c 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' + c 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}{2 a} e^{-\frac{b}{2 a} x} \right) + b e^{-\frac{b}{2 a} x} = \left( -b + b \right) e^{-\frac{b}{2 a} 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 never equal to zero we simply 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}} \neq 0.</math>
Thus <math>y_2(x)</math> is the second linearly independent solution we were looking for.
The General Method
Given a differential equation
- <math>y+p(t)y'+q(t)y=0\,</math>
and a single solution (<math>y_1(t)</math>), let the second solution be defined
- <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=0.</math>
Since <math>y_1(t)</math> is a solution of the original 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'=0</math>
which is a first-order differential equation for <math>v'(t)</math>. Divide by <math>y_1(t)</math>, obtaining
- <math>v+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=0</math>
and <math>v'(t)</math> can be found using a general method. Once <math>v'(t)</math> is solved, integrate it and enter into the original equation for <math>y_2</math>:
- <math>y_2=v(t)y_1(t).\,</math>
References
- 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.
- Eric W. Weisstein, Second-Order Ordinary Differential Equation Second Solution, From MathWorld--A Wolfram Web Resource.
