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 $$y_1(x)$$ 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 $$v$$.

An example
Consider the general, homogeneous, second-order linear constant coefficient ordinary differential equation. (ODE) $$ a y''(x) + b y'(x) + c y(x) = 0,$$ 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, $$b^2 - 4 a c$$, vanishes. In this case, $$ a y''(x) + b y'(x) + \frac{b^2}{4a} y(x) = 0,$$ 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 $$ a \left( v y_1 + 2 v' y_1' + v y_1 \right) + b \left( v' y_1 + v y_1' \right) + \frac{b^2}{4a} v y_1 = 0.$$ Rearranging this equation in terms of the derivatives of $$v(x)$$ we get $$ \left(a y_1 \right) v + \left( 2 a y_1' + b y_1 \right) v' + \left( a y_1 + b y_1' + \frac{b^2}{4a} y_1 \right) v = 0.$$

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

Therefore, we are left with $$ a y_1 v'' = 0.$$

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 $$ y_2(x) = ( c_1 x + c_2 ) y_1(x) = c_1 x y_1(x) + c_2 y_1(x).$$

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 $$ y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}.$$

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 $$W(y_1,y_2)(x) = \begin{vmatrix} y_1 & x y_1 \\ y_1' & y_1 + x y_1' \end{vmatrix} = y_1 ( y_1 + x y_1' ) - x y_1 y_1' = y_1^{2} + x y_1 y_1' - x y_1 y_1' = y_1^{2} = e^{-\frac{b}{a}x} \neq 0.$$

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 $$v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}.$$

One integrating factor is given by $$\mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}$$, and because

\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, $$ this integrating factor can be more neatly expressed as $$\mu(t) = e^{\ln(y_1^2(t)) + \int p(t)\,dt} = y_1^2(t)e^{\int p(t) dt}.$$ Multiplying the differential equation by the integrating factor $$\mu(t)$$, the equation for $$v(t)$$ can be reduced to $$\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, $$v'(t)$$ 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: $$y_2(t) = v(t)y_1(t).$$