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 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 straight forwardly found using characteristic equations except for the case when the discriminant, $b^2 - 4 a c$, vanishes. In this case,

$a y''(x) + b y'(x) + \frac{b^2}{4a} y(x) = 0, \;$

from which only one solution,

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

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

$a \left( v'' y_1 + 2 v' y_1' + v y_1'' \right) + b \left( v' y_1 + v y_1' \right) + \frac{b^2}{4a} v y_1 = 0.$

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

$\left(a y_1 \right) v'' + \left( 2 a y_1' + b y_1 \right) v' + \left( a y_1'' + b y_1' + \frac{b^2}{4a} y_1 \right) v = 0.$

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

• Variation of parameters

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.
• Teschl, Gerald (2012). Ordinary Differential Equations and Dynamical Systems. Providence: American Mathematical Society. ISBN 978-0-8218-8328-0.
• Eric W. Weisstein, Second-Order Ordinary Differential Equation Second Solution, From MathWorld—A Wolfram Web Resource.