Reduction of the 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 an (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)

$ay''(x)+by'(x)+cy(x)=0,$

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, $b^{2}-4ac$, vanishes. In this case,

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

from which only one solution,

$y_{1}(x)=e^{-{\frac {b}{2a}}x},$

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

$a\left(v''y_{1}+2v'y_{1}'+vy_{1}''\right)+b\left(v'y_{1}+vy_{1}'\right)+{\frac {b^{2}}{4a}}vy_{1}=0.$

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

$\left(ay_{1}\right)v''+\left(2ay_{1}'+by_{1}\right)v'+\left(ay_{1}''+by_{1}'+{\frac {b^{2}}{4a}}y_{1}\right)v=0.$

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)

$2a\left(-{\frac {b}{2a}}e^{-{\frac {b}{2a}}x}\right)+be^{-{\frac {b}{2a}}x}=\left(-b+b\right)e^{-{\frac {b}{2a}}x}=0.$

Therefore, we are left with

$ay_{1}v''=0.$

Since $a$ is assumed non-zero and $y_{1}(x)$ is an exponential function (and thus always non-zero), we 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

$y_{2}(x)=(c_{1}x+c_{2})y_{1}(x)=c_{1}xy_{1}(x)+c_{2}y_{1}(x).$

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

$y_{2}(x)=xy_{1}(x)=xe^{-{\frac {b}{2a}}x}.$

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

$W(y_{1},y_{2})(x)={\begin{vmatrix}y_{1}&xy_{1}\\y_{1}'&y_{1}+xy_{1}'\end{vmatrix}}=y_{1}(y_{1}+xy_{1}')-xy_{1}y_{1}'=y_{1}^{2}+xy_{1}y_{1}'-xy_{1}y_{1}'=y_{1}^{2}=e^{-{\frac {b}{a}}x}\neq 0.$

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

$v''+\left({\frac {2y_{1}'(t)}{y_{1}(t)}}+p(t)\right)\,v'={\frac {r(t)}{y_{1}(t)}}.$

The integrating factor is $\mu (t)=e^{\int ({\frac {2y_{1}'(t)}{y_{1}(t)}}+p(t))dt}=y_{1}^{2}(t)e^{\int p(t)dt}$.

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

${\frac {d}{dt}}\left(v'(t)y_{1}^{2}(t)e^{\int p(t)dt}\right)=y_{1}(t)r(t)e^{\int p(t)dt}.$

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:

$y_{2}(t)=v(t)y_{1}(t).$

Resources

• Reduction of the Order Notes. Produced by Paul Dawkins, Lamar University

References

1. 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.
2. Teschl, Gerald (2012). Ordinary Differential Equations and Dynamical Systems. Providence: American Mathematical Society. ISBN 978-0-8218-8328-0.
3. Eric W. Weisstein, Second-Order Ordinary Differential Equation Second Solution, From MathWorld—A Wolfram Web Resource.