An example

Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE)

${\displaystyle ay''(x)+by'(x)+cy(x)=0,}$

where ${\displaystyle a,b,c}$ 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, ${\displaystyle b^{2}-4ac}$, vanishes. In this case,

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

from which only one solution,

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

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

${\displaystyle y_{2}(x)=v(x)y_{1}(x)}$

where ${\displaystyle v(x)}$ is an unknown function to be determined. Since ${\displaystyle y_{2}(x)}$ must satisfy the original ODE, we substitute it back in to get

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

Rearranging this equation in terms of the derivatives of ${\displaystyle v(x)}$ we get

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

Since we know that ${\displaystyle y_{1}(x)}$ is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting ${\displaystyle y_{1}(x)}$ into the second term's coefficient yields (for that coefficient)

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

Therefore, we are left with

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

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

${\displaystyle v''=0.}$

This can be integrated twice to yield

${\displaystyle v(x)=c_{1}x+c_{2}}$

where ${\displaystyle c_{1},c_{2}}$ are constants of integration. We now can write our second solution as

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

Since the second term in ${\displaystyle y_{2}(x)}$ is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of

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

Finally, we can prove that the second solution ${\displaystyle y_{2}(x)}$ found via this method is linearly independent of the first solution by calculating the Wronskian

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

Thus ${\displaystyle y_{2}(x)}$ is the second linearly independent solution we were looking for.

General method

Given the general non-homogeneous linear differential equation

${\displaystyle y''+p(t)y'+q(t)y=r(t)}$

and a single solution ${\displaystyle y_{1}(t)}$ of the homogeneous equation [${\displaystyle r(t)=0}$], let us try a solution of the full non-homogeneous equation in the form:

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

where ${\displaystyle v(t)}$ is an arbitrary function. Thus

${\displaystyle y_{2}'=v'(t)y_{1}(t)+v(t)y_{1}'(t)}$

and

${\displaystyle y_{2}''=v''(t)y_{1}(t)+2v'(t)y_{1}'(t)+v(t)y_{1}''(t).}$

If these are substituted for ${\displaystyle y}$, ${\displaystyle y'}$, and ${\displaystyle y''}$ in the differential equation, then

${\displaystyle y_{1}(t)\,v''+(2y_{1}'(t)+p(t)y_{1}(t))\,v'+(y_{1}''(t)+p(t)y_{1}'(t)+q(t)y_{1}(t))\,v=r(t).}$

Since ${\displaystyle y_{1}(t)}$ is a solution of the original homogeneous differential equation, ${\displaystyle y_{1}''(t)+p(t)y_{1}'(t)+q(t)y_{1}(t)=0}$, so we can reduce to

${\displaystyle y_{1}(t)\,v''+(2y_{1}'(t)+p(t)y_{1}(t))\,v'=r(t)}$

which is a first-order differential equation for ${\displaystyle v'(t)}$ (reduction of order). Divide by ${\displaystyle y_{1}(t)}$, obtaining

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

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

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

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

After integrating the last equation, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): v'(t) is found, containing one constant of integration. Then, integrate ${\displaystyle v'(t)}$ to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:

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