Technique for solving linear ordinary differential equations
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
is known and a second linearly independent solution
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
.
Second-order linear ordinary differential equations[edit]
An example[edit]
Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE)
![{\displaystyle ay''(x)+by'(x)+cy(x)=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75c28f707e190e055decb44d7d0ddced643c8549)
where
![{\displaystyle a,b,c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f13f068df656c1b1911ae9f81628c49a6181194d)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc2c88a48e0087a5786b460b2e856d118b5e23ab)
, vanishes. In this case,
![{\displaystyle ay''(x)+by'(x)+{\frac {b^{2}}{4a}}y(x)=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69561bd9b1618e45d0c03301a966c4ce4af2bd04)
from which only one solution,
![{\displaystyle y_{1}(x)=e^{-{\frac {b}{2a}}x},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bc2de486bb26f707ef7a9a8c4af5ad5f6a8c2cf)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3306f5dd6d27e474007d0c239101d2ebf894f014)
where
![{\displaystyle v(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b371a381e15c71d8fc4ec43cf14b156f02a0d35a)
is an unknown function to be determined. Since
![{\displaystyle y_{2}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cccdc14b21f247e89df1c11787463b6e86d4289)
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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/286772aee7d28b3a0943e3efb2efeefa35db13b3)
Rearranging this equation in terms of the derivatives of
![{\displaystyle v(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b371a381e15c71d8fc4ec43cf14b156f02a0d35a)
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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e0519ca16de6b33fd199df521823d15b39e6426)
Since we know that
is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting
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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bc18f61a859e4e32ed5b2cceea85eaa3dfa343c)
Therefore, we are left with
![{\displaystyle ay_{1}v''=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/344a39625f1b5bcfaab6097e07514f2732af8fa6)
Since
is assumed non-zero and
is an exponential function (and thus always non-zero), we have
![{\displaystyle v''=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9defa24b3a91482c3103b5083de2aeafbea4922f)
This can be integrated twice to yield
![{\displaystyle v(x)=c_{1}x+c_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7494027e4a915c0fa91893225d6802c84e992cf1)
where
![{\displaystyle c_{1},c_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15ae7f7452e416f7b70940ca260b2582f10382cf)
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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c2c19430434c1dc1c005cb07f782638cb926b85)
Since the second term in
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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddff71484668eb8e6f30c0129464b8276e1d3241)
Finally, we can prove that the second solution
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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8804e22e6b590ff02f5c8e8d8d712a964773e9b)
Thus
is the second linearly independent solution we were looking for.
General method[edit]
Given the general non-homogeneous linear differential equation
![{\displaystyle y''+p(t)y'+q(t)y=r(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/881b7193abd92793dc19e47d2748e8df8cd5dc3b)
and a single solution
![{\displaystyle y_{1}(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b17efec23e0d4f93bac22db3e978eae43f9e9a0)
of the homogeneous equation [
![{\displaystyle r(t)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9042ffb812b65115649d4d286dc5366d1488d30)
], let us try a solution of the full non-homogeneous equation in the form:
![{\displaystyle y_{2}=v(t)y_{1}(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c5794e81cdcf92093cbf707712aef5299d2a36d)
where
![{\displaystyle v(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/243a0bf98a12f48552ba6a70302122d81b237b3d)
is an arbitrary function. Thus
![{\displaystyle y_{2}'=v'(t)y_{1}(t)+v(t)y_{1}'(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e5f6ed2f40b0052b67365eea65501d76004a92a)
and
![{\displaystyle y_{2}''=v''(t)y_{1}(t)+2v'(t)y_{1}'(t)+v(t)y_{1}''(t).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a945244aaedaae13768a3a7a9c94196097e10c3d)
If these are substituted for
,
, and
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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1544fe19da5a78be7f36856ce7fa32f9ede923d)
Since
is a solution of the original homogeneous differential equation,
, so we can reduce to
![{\displaystyle y_{1}(t)\,v''+(2y_{1}'(t)+p(t)y_{1}(t))\,v'=r(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23a2a850ea943607959e260f40977a6f6a37252e)
which is a first-order differential equation for
![{\displaystyle v'(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f8a024ea25190c68ccbcf96ccef94f9dc38b475)
(reduction of order). Divide by
![{\displaystyle y_{1}(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b17efec23e0d4f93bac22db3e978eae43f9e9a0)
, obtaining
![{\displaystyle v''+\left({\frac {2y_{1}'(t)}{y_{1}(t)}}+p(t)\right)\,v'={\frac {r(t)}{y_{1}(t)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e59a4dd29917afa4aafb9f57cddd607c1ca95ee2)
The integrating factor is
.
Multiplying the differential equation by the integrating factor
, the equation for
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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1902ec4c0fbb2e7c7b6b13adbc2e10e3110555a1)
After integrating the last equation,
is found, containing one constant of integration. Then, integrate
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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb9d2e0b40586933969afe6dfd550ebce1c27f08)
See also[edit]
References[edit]