Reduction of order

Reduction of order (or d’Alembert reduction) is a technique in mathematics for solving second-order linear ordinary differential equations. It is employed when one solution ${\displaystyle y_{1}(x)}$ is known and a second linearly independent solution ${\displaystyle y_{2}(x)}$ 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 ${\displaystyle v}$.

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.

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

One integrating factor is given by ${\displaystyle \mu (t)=e^{\int ({\frac {2y_{1}'(t)}{y_{1}(t)}}+p(t))dt}}$, and because

${\displaystyle \int \left({\frac {2y_{1}'(t)}{y_{1}(t)}}+p(t)\right)\,dt=2\int {\frac {y_{1}'(t)}{y_{1}(t)}}\,dt+\int p(t)\,dt=2\ln(y_{1}(t))+\int p(t)\,dt=\ln(y_{1}^{2}(t))+\int p(t)\,dt,}$

this integrating factor can be more neatly expressed as ${\displaystyle \mu (t)=e^{\ln(y_{1}^{2}(t))+\int p(t)\,dt}=y_{1}^{2}(t)e^{\int p(t)dt}.}$ 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, ${\displaystyle v'(t)}$ 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: $${\displaystyle y_{2}(t)=v(t)y_{1}(t).}$$

• Variation of parameters