# 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 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

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

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:

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