Reduction of order is a technique in mathematics for solving second-order linear 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. The method also applies to n-th order equations. In this case the ansatz will yield an (n-1)-th order equation for v .
Consider the general homogeneous second-order linear constant coefficient ordinary differential equation (ODE)
a y ″ ( x ) + b y ′ ( x ) + c y ( x ) = 0 , where a , b , c 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, b 2 − 4 a c , vanishes. In this case,
a y ″ ( x ) + b y ′ ( x ) + b 2 4 a y ( x ) = 0 ,
from which only one solution,
y 1 ( x ) = e − b 2 a x , 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
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 ( v ″ y 1 + 2 v ′ y 1 ′ + v y 1 ″ ) + b ( v ′ y 1 + v y 1 ′ ) + b 2 4 a v y 1 = 0. Rearranging this equation in terms of the derivatives of v ( x ) we get
( a y 1 ) v ″ + ( 2 a y 1 ′ + b y 1 ) v ′ + ( a y 1 ″ + b y 1 ′ + b 2 4 a y 1 ) 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 ( − b 2 a e − b 2 a x ) + b e − b 2 a x = ( − b + b ) e − 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 − 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 ) = | y 1 x y 1 y 1 ′ y 1 + x y 1 ′ | = 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 − b a x ≠ 0. Thus y 2 ( x ) is the second linearly independent solution we were looking for.
Given the general non-homogeneous linear differential equation
y ″ + p ( t ) y ′ + q ( t ) y = r ( t ) and a single solution y 1 ( t ) of the homogeneous equation [ r ( t ) = 0 ], let us try a solution of the full non-homogeneous equation in the form:
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 ) + 2 v ′ ( 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 ″ + ( 2 y 1 ′ ( t ) + p ( t ) y 1 ( t ) ) v ′ + ( y 1 ″ ( t ) + p ( t ) y 1 ′ ( t ) + q ( t ) y 1 ( t ) ) v = r ( t ) . Since y 1 ( t ) is a solution of the original homogeneous 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 ″ + ( 2 y 1 ′ ( t ) + p ( t ) y 1 ( t ) ) v ′ = r ( t ) which is a first-order differential equation for v ′ ( t ) (reduction of order). Divide by y 1 ( t ) , obtaining
v ″ + ( 2 y 1 ′ ( t ) y 1 ( t ) + p ( t ) ) v ′ = r ( t ) y 1 ( t ) .
Integrating factor: μ ( t ) = e ∫ ( 2 y 1 ′ ( t ) y 1 ( t ) + p ( t ) ) d t = y 1 2 ( t ) e ∫ p ( t ) d t .
Multiplying the differential equation with the integrating factor μ ( t ) , the equation for v ( t ) can be reduced to
d d t ( v ′ ( t ) y 1 2 ( t ) e ∫ p ( t ) d t ) = y 1 ( t ) r ( t ) e ∫ p ( t ) d t .
After integrating the last equation, v ′ ( t ) is found, containing one constant of integration. Then, integrate v ′ ( t ) to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:
y 2 ( t ) = v ( t ) y 1 ( t ) .
