Reduction of order

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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 y1(x) is known and a second linearly independent solution y2(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, b2 − 4ac, vanishes). Thus we have

\lambda_{1,2} = -\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 y2(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 y1(x) is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting y1(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 y1(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 c1,c2 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 y2(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 y2(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}} \neq 0.

Thus y2(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 (y1(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 y1(t) is a solution of the original differential equation, y1''(t) + p(t)y1'(t) + q(t)y1(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 y1(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 y2:

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

[edit] References

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