Reduction of order

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 a (n-1)-th order equation for v.

Second-order linear ordinary differential equations

An Example

Consider the general homogeneous second-order linear 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 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 \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.

General method

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)+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=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''+(2y_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''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}.

Integrating factor: \mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt}.

Multiplying the differential equation with the integrating factor \mu(t), the equation for v(t) can be reduced to

\frac{d}{dt}(v'(t) y_1^2(t) e^{\int p(t) dt})=y_1(t)r(t)e^{\int p(t) dt}.

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)\, .

See also

References