Reduction of order

An example

Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE) a y ″ ( x ) + b y ′ ( x ) + c y ( x ) = 0 , {\displaystyle ay''(x)+by'(x)+cy(x)=0,} where a , b , c {\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, b 2 − 4 a c {\displaystyle b^{2}-4ac} , vanishes. In this case, a y ″ ( x ) + b y ′ ( x ) + b 2 4 a y ( x ) = 0 , {\displaystyle ay''(x)+by'(x)+{\frac {b^{2}}{4a}}y(x)=0,} from which only one solution, y 1 ( x ) = e − b 2 a x , {\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 y 2 ( x ) = v ( x ) y 1 ( x ) {\displaystyle y_{2}(x)=v(x)y_{1}(x)} where v ( x ) {\displaystyle v(x)} is an unknown function to be determined. Since y 2 ( x ) {\displaystyle y_{2}(x)} must satisfy the original ODE, we substitute it back in to get a ( v ″ y 1 + 2 v ′ y 1 ′ + v y 1 ″ ) + b ( v ′ y 1 + v y 1 ′ ) + b 2 4 a v y 1 = 0. {\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 v ( x ) {\displaystyle v(x)} we get ( a y 1 ) v ″ + ( 2 a y 1 ′ + b y 1 ) v ′ + ( a y 1 ″ + b y 1 ′ + b 2 4 a y 1 ) v = 0. {\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 y 1 ( x ) {\displaystyle 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 ) {\displaystyle y_{1}(x)} into the second term's coefficient yields (for that coefficient) 2 a ( − b 2 a e − b 2 a x ) + b e − b 2 a x = ( − b + b ) e − b 2 a x = 0. {\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 a y 1 v ″ = 0. {\displaystyle ay_{1}v''=0.}

Since a {\displaystyle a} is assumed non-zero and y 1 ( x ) {\displaystyle y_{1}(x)} is an exponential function (and thus always non-zero), we have v ″ = 0. {\displaystyle v''=0.}

This can be integrated twice to yield v ( x ) = c 1 x + c 2 {\displaystyle v(x)=c_{1}x+c_{2}} where c 1 , c 2 {\displaystyle 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 x y 1 ( x ) + c 2 y 1 ( x ) . {\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 y 2 ( x ) {\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 y 2 ( x ) = x y 1 ( x ) = x e − b 2 a x . {\displaystyle y_{2}(x)=xy_{1}(x)=xe^{-{\frac {b}{2a}}x}.}

Finally, we can prove that the second solution y 2 ( x ) {\displaystyle y_{2}(x)} found via this method is linearly independent of the first solution by calculating the Wronskian W ( y 1 , y 2 ) ( x ) = | y 1 x y 1 y 1 ′ y 1 + x y 1 ′ | = y 1 ( y 1 + x y 1 ′ ) − x y 1 y 1 ′ = y 1 2 + x y 1 y 1 ′ − x y 1 y 1 ′ = y 1 2 = e − b a x ≠ 0. {\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 y 2 ( x ) {\displaystyle 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 ) {\displaystyle y''+p(t)y'+q(t)y=r(t)} and a single solution y 1 ( t ) {\displaystyle y_{1}(t)} of the homogeneous equation [ r ( t ) = 0 {\displaystyle r(t)=0} ], let us try a solution of the full non-homogeneous equation in the form: y 2 = v ( t ) y 1 ( t ) {\displaystyle y_{2}=v(t)y_{1}(t)} where v ( t ) {\displaystyle v(t)} is an arbitrary function. Thus y 2 ′ = v ′ ( t ) y 1 ( t ) + v ( t ) y 1 ′ ( t ) {\displaystyle y_{2}'=v'(t)y_{1}(t)+v(t)y_{1}'(t)} and y 2 ″ = v ″ ( t ) y 1 ( t ) + 2 v ′ ( t ) y 1 ′ ( t ) + v ( t ) y 1 ″ ( t ) . {\displaystyle y_{2}''=v''(t)y_{1}(t)+2v'(t)y_{1}'(t)+v(t)y_{1}''(t).}

If these are substituted for y {\displaystyle y} , y ′ {\displaystyle y'} , and y ″ {\displaystyle y''} in the differential equation, then y 1 ( t ) v ″ + ( 2 y 1 ′ ( t ) + p ( t ) y 1 ( t ) ) v ′ + ( y 1 ″ ( t ) + p ( t ) y 1 ′ ( t ) + q ( t ) y 1 ( t ) ) v = r ( t ) . {\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 y 1 ( t ) {\displaystyle 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 {\displaystyle y_{1}''(t)+p(t)y_{1}'(t)+q(t)y_{1}(t)=0} , so we can reduce to y 1 ( t ) v ″ + ( 2 y 1 ′ ( t ) + p ( t ) y 1 ( t ) ) v ′ = r ( t ) {\displaystyle 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 ) {\displaystyle v'(t)} (reduction of order). Divide by y 1 ( t ) {\displaystyle y_{1}(t)} , obtaining v ″ + ( 2 y 1 ′ ( t ) y 1 ( t ) + p ( t ) ) v ′ = r ( t ) y 1 ( t ) . {\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 μ ( t ) = e ∫ ( 2 y 1 ′ ( t ) y 1 ( t ) + p ( t ) ) d t {\displaystyle \mu (t)=e^{\int ({\frac {2y_{1}'(t)}{y_{1}(t)}}+p(t))dt}} , and because

∫ ( 2 y 1 ′ ( t ) y 1 ( t ) + p ( t ) ) d t = 2 ∫ y 1 ′ ( t ) y 1 ( t ) d t + ∫ p ( t ) d t = 2 ln ⁡ ( y 1 ( t ) ) + ∫ p ( t ) d t = ln ⁡ ( y 1 2 ( t ) ) + ∫ p ( t ) d t , {\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 μ ( t ) = e ln ⁡ ( y 1 2 ( t ) ) + ∫ p ( t ) d t = y 1 2 ( t ) e ∫ p ( t ) d t . {\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 μ ( t ) {\displaystyle \mu (t)} , the equation for v ( t ) {\displaystyle v(t)} can be reduced to d d t ( v ′ ( t ) y 1 2 ( t ) e ∫ p ( t ) d t ) = y 1 ( t ) r ( t ) e ∫ p ( t ) d t . {\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, v ′ ( t ) {\displaystyle v'(t)} is found, containing one constant of integration. Then, integrate v ′ ( t ) {\displaystyle 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 ) . {\displaystyle y_{2}(t)=v(t)y_{1}(t).}