# Preface

This section presents a technique how to find another linearly independent solution to the linear differential equation when one solution is known.

# Reduction of order

We recall the definition of a root multiplicity. A real or complex number α is called a root of multiplicity k of the polynomial p(x) if there is a polynomial s(x) such that $$s(\alpha ) \ne 0$$ and $$p(x) = \left( x- \alpha \right)^k s(x) .$$ If k=1, then α is called a simple root. If k ≥ 2, then α is called a multiple root.

Reduction of order for constant coefficient equations

$$\label{EqReduction.1} a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0 , \qquad a_n \ne 0.$$
Here 𝑎0, 𝑎1, … , 𝑎n are real constants. It is convenient to assign a linear differential operator to this equation
$$\label{EqReduction.2} L \left[ \texttt{D} \right] = a_n \texttt{D}^n + a_{n-1} \texttt{D}^{n-1} + \cdots + a_1 \texttt{D} + a_0 \texttt{D}^0 , \qquad \texttt{D} = {\text d}/{\text d}x ,$$
where $$\texttt{D}^0$$ is the identity operator, denoted by I. Then Eq.\eqref{EqReduction.1} can be rewritten in compact form
$$\label{EqReduction.3} L \left[ \texttt{D} \right] y = 0 .$$
Suppose that the characteristic polynomial $$L \left( \lambda \right) = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0 ,$$ corresponding to the operator L, has a repeated root r of multiplicity m. That is, the characteristic polynomial can be factored $$L \left( \lambda \right) = \left( \lambda - r \right)^m M(\lambda ) ,$$ where M(λ) is a polynomial of degree n - m such that M(r) ≠ 0. For example, the second order constant coefficient linear differential operator $$L \left[ \texttt{D} \right] = a\,\texttt{D}^2 + b\,\texttt{D} +c$$ has a repeated factor if and only if the corresponding characteristic equation $$a\,\lambda^2 + b\,\lambda + c =0$$ has a double root:
$\lambda_1 = \lambda_2 = -b/(2a) .$
In other words, the quadratic polynomial can be factored $$a\,\lambda^2 + b\,\lambda + c = a \left( \lambda - \lambda_1 \right)^2$$ if and only if its discriminant $$b^2 - 4ac$$ is zero. In this case we have only one solution of exponential form:
$$\label{EqReduction.4} y_1 (x) = e^{-bx/(2a)} = e^{\lambda_1 x}$$
for the differential equation
$$\label{EqReduction.5} a\, y'' + b\, y' + c\,y =0, \qquad b^2 = 4ac.$$

To find another linearly independent solution to the homogeneous equation \eqref{EqReduction.5} , we use the method of reduction of order, credited to Jacob Bernoulli (1655--1705). Setting

$$\label{EqReduction.6} y(x) = v(x)\, y_1 (x) = v(x)\, e^{-bx/(2a)} ,$$
we have
\begin{align*} y' &= v' (x)\, y_1 (x) + v(x)\, y'_1 (x) = \left[ v' (x) - \frac{b}{2a}\, v(x) \right] y_1 (x) , \\ y'' &= \left[ v'' (x) - \frac{b}{a}\, v' (x) + \left( \frac{b}{2a} \right)^2 \right] y_1 (x) . \end{align*}
Upon substituting the Bernoulli form \eqref{EqReduction.6} into the differential equation \eqref{EqReduction.5}, we get
\begin{align*} L \left[ \texttt{D} \right] y &= a \left[ v'' (x) - \frac{b}{a}\, v' (x) + \left( \frac{b}{2a} \right)^2 \right] y_1 (x) + b \left[ v' (x) - \frac{b}{2a}\, v(x) \right] y_1 (x) + c\, v(x)\, y_1 (x) \\ &= a\, v'' (x)\, y_1 (x) + v(x) \,L \left[ \texttt{D} \right] y_1 (x) = a\, v'' (x)\, y_1 (x) =0. \end{align*}
The latter can be divided by nonzero term $$a\,y_1 (x) = a\,e^{-bx/(2a)} .$$ This yields $$v'' (x) =0,$$ and after integration, we obtain $$v (x) = C_1 + C_2 x,$$ with arbitrary constants C1 and C2. Finally, substituting for v(x), we define the general solution
$y(x) = v(x)\, y_1 (x) = \left( C_1 + C_2 x \right) e^{-bx/(2a)} .$
Functions $$y_1 (x) = e^{-bx/(2a)} \quad \mbox{and} \quad y_2 (x) = x\,e^{-bx/(2a)}$$ are linearly independent because their Wronskian
$W(x) = \begin{vmatrix} e^{\gamma x} & x\, e^{\gamma x} \\ \gamma \, e^{\gamma x} & \left( x\,\gamma + 1 \right) e^{\gamma x} \end{vmatrix} = e^{2\gamma x} \ne 0 .$
Therefore, these two functions form a fundamental set of solutions whenever γ is.     ■

This method can be extended for arbitrary case when a linear constant coefficient differential operator has a multiple root $$L \left[ \texttt{D} \right] = \left( \texttt{D} - \gamma \right)^m M \left[ \texttt{D} \right]$$ of multiplicity m. Then to such multiple correspond m linearly independent solutions:

$y_1 (x) = e^{-x\gamma} , \ y_2 (x) = x\, e^{-x\gamma} , \ \ldots , \ y_m (x) = x^{m-1} e^{-x\gamma} .$
The strict derivation of the above statement is made in annihilation section.

Theorem 1: Let $$a_0 , a_1 , \ldots , a_n$$ be n+1 real (or complex) numbers with 𝑎n ≠ 0, and y(x) be a n times continuously differentiable function on some interval | 𝑎, b|. Then y(x) is a solution of the n-th order linear differential equation with constant coefficients

$L \left[ \texttt{D} \right] y \equiv a_n \texttt{D}^n y + a_{n-1} \texttt{D}^{n-1} y + \cdots + a_1 \texttt{D} \, y+ a_0 \, y =0 , \qquad \texttt{D} = {\text d}/{\text d} x ,$
if and only if
$y (x) = \sum_{j=1}^m e^{-x\lambda_j x} \, P_j (x) ,$
where $$\lambda_1 , \lambda_2 , \ldots , \lambda_m$$ are distinct roots of the characteristic polynomial $$\sum_{k=1}^n a_k \lambda^k$$ with multiplicities $$m_1 , m_2 , \ldots , m_m ,$$ respectively, and $$P_k (x)$$ is a polynomial of degree $$m_k -1 .$$       ⧫

For illustration, we consider a particular case of second order differential equation---the motion for undriven damped harmonic oscillator (coefficients μ and ω0 > 0 are assumed constants):

$\ddot{y} + 2\mu\,\dot{y} + \omega_0^2 y =0 ,$
subject to the initial conditions
$y(0) = y_0 , \qquad \dot{y}(0) = v_0 .$
We show that its solution can be obtained from the solution
\begin{align*} y(t) &= \frac{y_0 \lambda_2 - v_0}{\lambda_2 - \lambda_1}\, e^{\lambda_1 t} - \frac{y_0 \lambda_1 - v_0}{\lambda_2 - \lambda_1}\, e^{\lambda_2 t} \\ &= y_0 \, e^{\lambda_1 t} + \left( y_0 \lambda_1 - v_0 \right) \left[ \frac{e^{\lambda_1 t} - e^{\lambda_2 t}}{\lambda_2 - \lambda_1} \right] \end{align*}
by taking the limit when $$\lambda_1 \ \to \, \lambda_2 .$$ Here
$\lambda_1 , \ \lambda_2 = - \mu \pm \eta = -\mu \pm \left( \mu^2 - \omega_0^2 \right)^{1/2} , \qquad \eta = \left( \mu^2 - \omega_0^2 \right)^{1/2}$
are two distinct roots of the characteristic equation $$\lambda^2 + 2\mu\,\lambda + \omega_0^2 =0 .$$ Let assume, for simplicity, that $$\eta = \left( \mu^2 - \omega_0^2 \right)^{1/2} \ge 0 ,$$ that is $$\mu \ge \omega_0 .$$ Then rewriting the solution in the equivalent form:
$y(t) = y_0 e^{-\mu\, t} \,\cosh (\eta \,t) + \left( \mu y_0 + v_0 \right) e^{-\mu\,t} \,\frac{\sinh (\eta\, t)}{\eta} , \qquad \eta = \left( \mu^2 - \omega_0^2 \right)^{1/2} \ge 0,$
and taking the limit as $$\eta \, \to \, 0,$$ one gets
$y(t) = e^{-\mu\, t} \left[ y_0 + \left( \mu y_0 + v_0 \right) t \right] .$

Example: Let us consider the differential equation

$y'' -4\,y' +4\,y =0 .$
The characteristic equation $$\lambda^2 - 4\,\lambda +4 = \left( \lambda -2 \right)^2 =0$$ has a double root $$\lambda =2 .$$ Hence we get one exponential solution $$y_1 (x) = e^{2x}$$ and another one is a multiple of the latter: $$y_2 (x) = x\,e^{2x} .$$ So the general solution becomes
$y(x) = C_1 e^{2x} + C_2 x\,e^{2x} ,$
with some arbitrary constants C1 and C2.
DSolve[y''[x]-4 y'[x]+4 y[x]==0,y[x],x]
soln[x_] = Expand[y[x]/.%[[1]]]
y1[x_] = Coefficient[soln[x],C[1]]
y2[x_] = Coefficient[soln[x], C[2]]
basis = {y1[x], y2[x]}
WronskianDet = Det[{basis, D[basis, x]}]
Out[22]= E^(4 x)

■

Example: Consider the initial value problem

$y'' +6\,y' +9\,y =0 , \qquad y(0) = -1, \quad y' =(0) = 5.$
The characteristic polynomial $$\lambda^2 + 6\,\lambda +9 = = \left( \lambda +3 \right)^2$$ has one double root $$\lambda =-3 .$$ Therefore, the general solution of the given homogeneous differential equation becomes
$y(x) = C_1 e^{-3x} + C_2 x\, e^{-3x} .$
To satisfy the initial conditions, we have to choose arbitrary constants that are solutions of the system of equations:
$\begin{split} y(0) = C_1 =-1, \\ y'(0) = -3\, C_1 + C_2 =5. \end{split}$
So $$C_1 =-1, \quad C_2 =2$$ and the required solution becomes
$y(x) = - e^{-3x} + 2\, x\, e^{-3x} .$

soln = DSolve[{x''[t] + 6 x'[t] + 9 x[t] == 0, x[0] == -1,
x'[0] == 5}, x[t], t]
s2[t_] = x[t] /. soln[[1]]
Plot[s2[t], {t, 0.2, 3}]

Out[11]= {{x[t] -> E^(-3 t) (-1 + 2 t)}}
Out[12]= E^(-3 t) (-1 + 2 t)
Out[13]=

When does the extreme occur?
Solve[s2'[t] == 0, t]
out[48]= {{t -> 5/6}}

Where is the extreme?
s2[t /. %]
Out[49]= {2/(3 E^(5/2))}
N[%]
Out[50]= {0.0547233}

■

We generalized considered above two examples by introducing the homogeneous differential equation
$y'' -2\alpha\,y' + \alpha^2 y =0 ,$
which has a solution $$y(x) = e^{\alpha\, x} .$$ Letting, $$y(x) = e^{\alpha\, x} u(x) ,$$ we obtain its derivatives $$y' =\alpha\, y + e^{\alpha\, x} u' (x) ,$$ and $$y'' =\alpha^2\, y + 2\,e^{\alpha\, x} u (x) + e^{\alpha\, x} u'' (x) .$$ Substituting into the equation and cancelling the common exponential multiple yields
$u'' -2\alpha\,u' + 2\alpha\,u' =0 \qquad \Longrightarrow \qquad u'' =0.$
ntegrating twice gives$$u = A + B\,x .$$ Hence, the second linearly independent solution becomes
$y(x) = x\, e^{\alpha\, x} .$

Example: Let us consider the fourth order linear differential operator:

$L \left[ \texttt{D} \right] y \equiv \texttt{D}^4 y - 8\, \texttt{D}^{3} y + 18\, \texttt{D}^2 \, y -27 \, y =0 , \qquad \texttt{D} = {\text d}/{\text d} x .$
The corresponding characteristic polynomial
$L \left( \lambda \right) = \lambda^4 - 8\, \lambda^{3} + 18\, \lambda^2 -27 = \left( \lambda -3 \right)^3 \left( \lambda +1 \right)$
has one simple root %lambda; = −1 ≠ 0 and a triple root λ = 3. Therefore, the general solution to the fourth order differential equation $$L \left[ \texttt{D} \right] y =0$$ is
$y(x) = C_1 e^{-x} + \left( C_2 + C_3 x + C_4 x^2 \right) e^{3x} ,$
where Ck, $$k=1,2,3,4 ,$$ are arbitrary constants.    ■

Variable coefficient equations

The reduction of order technique, which applies to arbitrary linear differential equations, allows us to go beyond equations with constant coefficients, provided that we already know one solution. For sake of clarity, we start with a second order linear differential equation with variable coefficients:

$$\label{EqReduction.7} a_2 (x)\, y'' + a_1 (x)\, y' + a_0 (x)\,y =0, \qquad a_2 (x) \ne 0,$$
where 𝑎0(x), 𝑎1(x), 𝑎2(x) are some continuous functions in an interval |α, β|. Let L be a corresponding differential operator
$$\label{EqReduction.8} L \left[ x, \texttt{D} \right] = a_2 (x)\, \texttt{D}^2 + a_1 (x)\, \texttt{D} + a_0 (x)\,\texttt{D}^0, \qquad \texttt{D} = {\text d}/{\text d}x .$$
Since $$\texttt{D}^0 = \texttt{I} , |0 the identity operator, the adjoint to operator \eqref{EqReduction.8} is written as $$\label{EqReduction.9} L^{\ast} \left[ x, \texttt{D} \right] = a_2 (x)\, \texttt{D}^2 + \left( 2\,a'_2 (x) - a_1 (x)\right) \texttt{D} + \left( a_0 (x) - a'_1 (x) + a''_2 (x) \right) \texttt{I}, \qquad \texttt{D}^0 = \texttt{I} .$$ Theorem 2: If φ(x) is a solution of a second order homogeneous equation $a_2 (x)\,\varphi'' + a_1 (x)\, \varphi' + a_0 (a) \,\varphi =0 ,$ then the function $\psi (x) = \frac{1}{a_2 (x)\,W[x]} \,\varphi (x)$ is a solution of the adjoint homogeneous equation $a_2 \psi'' + \left( 2\,a'_2 - a'_1 \right) \psi' + \left( a''_2 - a'_1 + a_0 \right) \psi = 0 .$ Here \( W[x] = W\left( x_0 \right) \exp \left\{ -\int a_1 (x)/a_2 (x)\,{\text d}x \right\}$$ is the Wronskian (arbitrary multiplier is ommited) of the operator \eqref{EqReduction.8}.    ⧫

Example 4: Consider the Chebyshev equation

$\left( 1 - x^2 \right) y'' -x\,y' +y =0 .$
It is easy to veryfy that this equation has a linear function solution φ(x) = x. However, it is very difficult to guess a solution for the adjoint differential equation
$\left( 1 - x^2 \right) w'' -3x\,w' =0 .$
Using Theorem 2, we obtain its solution
$\psi (x) = x \left( 1 - x^2 \right)^{-1/2}$
because
Integrate[x/(1 - x^2), x]
-(1/2) Log[1 - x^2]
$W[x] = \exp\left\{ \int \frac{x}{1-x^2}\,{\text d}x \right\} = \left( 1 - x^2 \right)^{-1/2} .$
We check with Mathematica
psi[x_] = x*(1 - x^2)^(-1/2); Simplify[D[psi[x], x, x]*(1 - x^2) - 3*x*D[psi[x], x] + 0*psi[x]]
0
■
To get rid of a trouble maker, the leading coefficient 𝑎2 in Eq.\eqref{EqReduction.7}, we divide every term in this equation to obtain the second order differential equation in normal form:
$y'' +p(x)\,y' +q(x)\,y =0 ,$
where p(x) and q(x) are some continuous functions on some interval |a,b|. Suppose that we know one its solution $$y = y_1 (x) \ne 0.$$ This means that
$L \left[ \texttt{D} \right] y_1 =0 , \qquad\mbox{where} \quad L \left[ \texttt{D} \right] = \texttt{D}^2 + p(x)\,\texttt{D} + q(x) .$
Then we seek another linearly independent solution in the form $$y(x) = v(x)\, y_1 (x) ,$$ where unknown function v(x) s determined upon substitution of this form into the given differential equation. First, we use the product rule to obtain
\begin{align*} y' &= v' (x)\, y_1 (x) + v(x)\, y'_1 (x) , \\ y'' &= v'' (x)\, y_1 (x) +2\, v' (x)\, y_1 (x) + v(x)\, y''_1 (x) . \end{align*}
This implies that we must have
\begin{align*} L \left[ \texttt{D} \right] v(x)\,y_1 &= v'' (x)\, y_1 (x) +2\, v' (x)\, y'_1 (x) + v(x)\, y''_1 (x) + p(x) \left[ v' (x)\, y_1 (x) + v(x)\, y'_1 (x) \right] + q(x)\,v(x)\,y_1 (x) \\ &= v'' (x)\, y_1 (x) +2\, v' (x)\, y'_1 (x) + p(x)\,v' (x)\, y_1 (x) + v(x)\, L \left[ \texttt{D} \right] y_1 =0. \end{align*}
Since $$L \left[ \texttt{D} \right] y_1 \equiv 0 ,$$ the function v(x) must satisfy the equation
$v'' (x)\, y_1 (x) + v' (x) \left[ 2\,y'_1 (x) + p(x) \, y_1 (x) \right] =0 .$
This is a second order differential equation where dependent function is missing. If we set $$u= v' ,$$ we reduce it to a first order differential equation:
$u' (x)\, y_1 (x) = - u (x) \left[ 2\,y'_1 (x) + p(x) \, y_1 (x) \right] \qquad\Longrightarrow \qquad \frac{{\text d} u}{u} = - \frac{1}{y_1 (x)} \left[ 2\,y'_1 (x) + p(x) \, y_1 (x) \right] {\text d} x .$
Integrating both sides, we get
$\ln u = - 2\int \frac{y'_1 (x)}{y_1 (x)} \,{\text d}x - \int p(x) \, {\text d} x = \ln y_1^{-2} - \int p(x) \, {\text d} x .$
This allows us to determine u(x) explicitly:
$v' = u = y_1^{-2} \, \exp \left\{ - \int p(x) \, {\text d} x \right\} ,$
where we dropped an arbitrary constant because we are after another linearly independent solution. Next integration yields
$v = \int y_1^{-2} \, \exp \left\{ - \int p(x) \, {\text d} x \right\} {\text d} x \qquad\Longrightarrow \qquad y_2 (x) = y_1 (x) \, \int y_1^{-2} \, \exp \left\{ - \int p(x) \, {\text d} x \right\} {\text d} x .$

Theorem 3: Let φ1(x be a solution of the second order homogeneous equation $$a_2 (x)\,y'' + a_1 (x)\,y' + a_0 (x)\,y =0 .$$ Then the second linearly independent solution to Eq.\eqref{EqReduction.7} is

$\varphi_2 (x) = \varphi_1 (x) \int \frac{W[x]}{\varphi_1^2 (x)}\,{\text d}x = a_2 W[x]\,\psi_1 (x) \int \frac{{\text d}x}{a_2^2 \psi_1^2 W[x]} ,$
where ψ1(x) is a solution of the adjoint homogeneous equation $$L^{\ast} \left[ x, \texttt{D} \right] \psi = 0 ,$$ and $$W[x] = \exp \left\{ -\int a_1 (x)/a_2 (x)\,{\text d}x \right\}$$ is the Wronskian (arbitrary multiplier is ommited) of the operator \eqref{EqReduction.8}.    ⧫

Example 5: Find a second linearly independent solution to the differential equation

$t\,y'' - y' + 4t^3\,y =0 , \qquad\mbox{for} \quad t>0,$
provided that $$y_1 (t) = \sin \left( t^2 \right)$$ is a solution. First, let's rewrite our differential equation by diving by t to get:
$y'' - \frac{1}{t}\,y' + 4t^2\,y =0 .$
We seek another solution in the form $$y= v(t)\, y_1 (t) = v(t)\, \sin \left( t^2 \right) .$$ Then for u=v'(t) we have the separable equation
$\frac{u'}{u} = - 2\,\frac{y'_1}{y_1} + \frac{1}{t} \qquad\Longrightarrow \qquad \ln u = \ln \left( t\,y_1^{-2} \right) .$
Then
$v' = u = \frac{t}{y_1} \qquad\Longrightarrow \qquad u = \int \frac{t}{\sin^2 t^2} \, {\text d}t = \frac{1}{2}\, \cot \left( t^2 \right) .$
So another linearly independent solution becomes
$y_2 (t) = v(t)\,y_1 (t) = \frac{1}{2}\,\cos \left( t^2 \right) .$
Of course, the constant multiple (1/2) could be dropped.    ■

1. One solution of $$x^2 y'' - \left( x^2 + 2x \right) y' + \left( x+2 \right) y = 0$$ is y(x) = x, Use the method of reduction of order to find a 2nd solution and hence write down the general solution.