Reduction of order

Reduction of order

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.

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_{1,2} = -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" ight) + b left( v' y_1 + v y_1' ight) + c v y_1 = 0.

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

: left(a y_1 ight) v" + left( 2 a y_1' + b y_1 ight) v' + left( a y_1" + b y_1' + c y_1 ight) 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} ight) + b e^{-frac{b}{2 a} x} = left( -b + b ight) 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) = egin{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 eq 0.

Thus y_2(x) is the second linearly independent solution we were looking for.

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) ight),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).,

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, " [http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquationSecondSolution.html Second-Order Ordinary Differential Equation Second Solution] ", From MathWorld--A Wolfram Web Resource.


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