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 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 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}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)}}

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)\,

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.

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Reduction of order

Second-order linear ordinary differential equations

An Example

General method

See also

References