Reduction of order

Short description: 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 [math]\displaystyle{ y_1(x) }[/math] is known and a second linearly independent solution [math]\displaystyle{ y_2(x) }[/math] 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 [math]\displaystyle{ v }[/math].

## Contents

* 1 Second-order linear ordinary differential equations
* 1.1 An example
* 1.2 General method
* 2 See also
* 3 References

## Second-order linear ordinary differential equations

### An example

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

[math]\displaystyle{ a y''(x) + b y'(x) + c y(x) = 0, }[/math]

where [math]\displaystyle{ a, b, c }[/math] 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, [math]\displaystyle{ b^2 - 4 a c }[/math], vanishes. In this case,

[math]\displaystyle{ a y''(x) + b y'(x) + \frac{b^2}{4a} y(x) = 0, }[/math]

from which only one solution,

[math]\displaystyle{ y_1(x) = e^{-\frac{b}{2a} x}, }[/math]

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

[math]\displaystyle{ y_2(x) = v(x) y_1(x) }[/math]

where [math]\displaystyle{ v(x) }[/math] is an unknown function to be determined. Since [math]\displaystyle{ y_2(x) }[/math] must satisfy the original ODE, we substitute it back in to get

[math]\displaystyle{ 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. }[/math]

Rearranging this equation in terms of the derivatives of [math]\displaystyle{ v(x) }[/math] we get

[math]\displaystyle{ \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. }[/math]

Since we know that [math]\displaystyle{ y_1(x) }[/math] is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting [math]\displaystyle{ y_1(x) }[/math] into the second term's coefficient yields (for that coefficient)

[math]\displaystyle{ 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. }[/math]

Therefore, we are left with

[math]\displaystyle{ a y_1 v'' = 0. }[/math]

Since [math]\displaystyle{ a }[/math] is assumed non-zero and [math]\displaystyle{ y_1(x) }[/math] is an exponential function (and thus always non-zero), we have

[math]\displaystyle{ v'' = 0. }[/math]

This can be integrated twice to yield

[math]\displaystyle{ v(x) = c_1 x + c_2 }[/math]

where [math]\displaystyle{ c_1, c_2 }[/math] are constants of integration. We now can write our second solution as

[math]\displaystyle{ y_2(x) = ( c_1 x + c_2 ) y_1(x) = c_1 x y_1(x) + c_2 y_1(x). }[/math]

Since the second term in [math]\displaystyle{ y_2(x) }[/math] is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of

[math]\displaystyle{ y_2(x) = x y_1(x) = x e^{-\frac{b}{2 a} x}. }[/math]

Finally, we can prove that the second solution [math]\displaystyle{ y_2(x) }[/math] found via this method is linearly independent of the first solution by calculating the Wronskian

[math]\displaystyle{ 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. }[/math]

Thus [math]\displaystyle{ y_2(x) }[/math] is the second linearly independent solution we were looking for.

### General method

Given the general non-homogeneous linear differential equation

[math]\displaystyle{ y'' + p(t)y' + q(t)y = r(t) }[/math]

and a single solution [math]\displaystyle{ y_1(t) }[/math] of the homogeneous equation [[math]\displaystyle{ r(t)=0 }[/math]], let us try a solution of the full non-homogeneous equation in the form:

[math]\displaystyle{ y_2 = v(t)y_1(t) }[/math]

where [math]\displaystyle{ v(t) }[/math] is an arbitrary function. Thus

[math]\displaystyle{ y_2' = v'(t)y_1(t) + v(t)y_1'(t) }[/math]

and

[math]\displaystyle{ y_2'' = v''(t)y_1(t) + 2v'(t)y_1'(t) + v(t)y_1''(t). }[/math]

If these are substituted for [math]\displaystyle{ y }[/math], [math]\displaystyle{ y' }[/math], and [math]\displaystyle{ y'' }[/math] in the differential equation, then

[math]\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). }[/math]

Since [math]\displaystyle{ y_1(t) }[/math] is a solution of the original homogeneous differential equation, [math]\displaystyle{ y_1''(t)+p(t)y_1'(t)+q(t)y_1(t)=0 }[/math], so we can reduce to

[math]\displaystyle{ y_1(t)\,v'' + (2y_1'(t)+p(t)y_1(t))\,v' = r(t) }[/math]

which is a first-order differential equation for [math]\displaystyle{ v'(t) }[/math] (reduction of order). Divide by [math]\displaystyle{ y_1(t) }[/math], obtaining

[math]\displaystyle{ v''+\left(\frac{2y_1'(t)}{y_1(t)}+p(t)\right)\,v'=\frac{r(t)}{y_1(t)}. }[/math]

The integrating factor is [math]\displaystyle{ \mu(t)=e^{\int(\frac{2y_1'(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t) dt} }[/math].

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

[math]\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}. }[/math]

After integrating the last equation, [math]\displaystyle{ v'(t) }[/math] is found, containing one constant of integration. Then, integrate [math]\displaystyle{ v'(t) }[/math] to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:

[math]\displaystyle{ y_2(t) = v(t)y_1(t). }[/math]

## See also

* Variation of parameters

## References

* W. E. Boyce and R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems (8th edition), John Wiley & Sons, Inc., 2005. ISBN:0-471-43338-1.
* Teschl, Gerald (2012). Ordinary Differential Equations and Dynamical Systems. Providence: American Mathematical Society. ISBN 978-0-8218-8328-0. https://www.mat.univie.ac.at/~gerald/ftp/book-ode/.
* Eric W. Weisstein, Second-Order Ordinary Differential Equation Second Solution, From MathWorld—A Wolfram Web Resource.

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