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Reduction of order is a technique in mathematics for solving second-order ordinary differential equations. It is employed when one solution $y 1 (x)$ is known and a second linearly independent solution $y 2 (x)$ is desired.

[edit] A Simple Example

Consider the general second-order constant coefficient ODE

$a y''(x) + b y'(x) + c y(x) = 0, \;$

where $a, b, c$ are real non-zero coefficients. Furthermore, assume that the associated characteristic equation

$a \lambda^{2} + b \lambda + c = 0 \;$

has repeated roots (i.e. the discriminant, $b 2 - 4 a c$ , vanishes). Thus we have

$\lambda_1 = \lambda_2 = -\frac{b}{2 a}.$

Thus our one solution to the ODE is

$y_1(x) = e^{-\frac{b}{2 a} x}.$

To find a second solution we take as an ansatz

$y_2(x) = v(x) y_1(x) \;$

where $v (x)$ is an unknown function to be determined. Since $y 2 (x)$ must satisfy the original ODE, we substitute it back in to get

$a \left( v'' y_1 + 2 v' y_1' + v y_1'' \right) + b \left( v' y_1 + v y_1' \right) + c v y_1 = 0.$

Rearranging this equation in terms of the derivatives of $v (x)$ we get

$\left(a y_1 \right) v'' + \left( 2 a y_1' + b y_1 \right) v' + \left( a y_1'' + b y_1' + c y_1 \right) v = 0.$

Since we know that $y 1 (x)$ is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting $y 1 (x)$ into the second term's coefficient yields (for that coefficient)

$2 a \left( - \frac{b}{2 a} e^{-\frac{b}{2 a} x} \right) + b e^{-\frac{b}{2 a} x} = \left( -b + b \right) e^{-\frac{b}{2 a} x} = 0.$

Therefore we are left with

$a y_1 v'' = 0. \;$

Since $a$ is assumed non-zero and $y 1 (x)$ is an exponential function and thus never equal to zero we simply have

$v'' = 0. \;$

This can be integrated twice to yield

$v(x) = c_1 x + c_2 \;$

where $c 1, c 2$ are constants of integration. We now can write our second solution as

$y_2(x) = ( c_1 x + c_2 ) y_1(x) = c_1 x y_1(x) + c_2 y_1(x). \;$

Since the second term in $y 2 (x)$ is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of

$y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}.$

Finally, we can prove that the second solution $y 2 (x)$ found via this method is linearly independent of the first solution by calculating the Wronskian

$W(y_1,y_2)(x) = \begin{vmatrix} y_1 & x y_1 \\ y_1' & y_1 + x y_1' \end{vmatrix} = y_1 ( y_1 + x y_1' ) - x y_1 y_1' = y_1^{2} + x y_1 y_1' - x y_1 y_1' = y_1^{2} = e^{-\frac{b}{a}x} \neq 0.$

Thus $y 2 (x)$ is the second linearly independent solution we were looking for.

[edit] The General Method

Given a differential equation

$y''+p(t)y'+q(t)y=0\,$

and a single solution ( $y 1 (t)$ ), let the second solution be defined

$y_2=v(t)y_1(t)\,$

where $v (t)$ is an arbitrary function. Thus

$y_2'=v'(t)y_1(t)+v(t)y_1'(t)\,$

and

$y_2''=v''(t)y_1(t)+2v'(t)y_1'(t)+v(t)y_1''(t).\,$

If these are substituted for $y$ , $y'$ , and $y''$ in the differential equation, then

$y_1(t)\,v''+(2y_1'(t)+p(t)y_1(t))\,v'+(y_1''(t)+p(t)y_1'(t)+q(t)y_1(t))\,v=0.$

Since $y 1 (t)$ is a solution of the original differential equation, $y 1''(t) + p (t) y 1'(t) + q (t) y 1 (t) = 0$ , so we can reduce to

$y_1(t)\,v''+(2y_1'(t)+p(t)y_1(t))\,v'=0$

which is a first-order differential equation for $v'(t)$ . Divide by $y 1 (t)$ , obtaining

$v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=0$

and $v'(t)$ can be found using a general method. Once $v'(t)$ is solved, integrate it and enter into the original equation for $y 2$ :

$y_2=v(t)y_1(t).\,$

[edit] 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.