Higher-Order ODEs Polynomial Form

Abstract

If a function $f(t,x,\dot{x},\ldots,x^{(n)})$ is defined and is continuous in a subset of $\mathbb{R}^{n+2}(n \geq 1)$, then the equation

$$\displaystyle{ f(t,x,\dot{x},\ldots,x^{(n)}) = 0 }$$

(5.1)

is said to be ordinary differential equation of nth-order.

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5.1 Introduction

If a function $f(t,x,\dot{x},\ldots,x^{(n)})$ is defined and is continuous in a subset of $\mathbb{R}^{n+2}(n \geq 1)$, then the equation

$$\displaystyle{ f(t,x,\dot{x},\ldots,x^{(n)}) = 0 }$$

(5.1)

is said to be ordinary differential equation of nth-order.

Fortunately, it happens very often that (5.1) can be transformed to the form

$$\displaystyle{ x^{(n)} = f_{ 1}(t,x,\dot{x},\ldots,x^{(n-1)}). }$$

(5.2)

The Cauchy problem for Eq. (5.2) is that of finding a solution x(t) satisfying the series of the following initial conditions

$$\displaystyle{ x(t_{0}) = x_{0},\ \ \dot{x}(t_{0}) = \dot{x}_{0},\ldots,x^{(n-1)}(t_{ 0}) = x_{0}^{(n-1)}. }$$

(5.3)

Theorem 5.1 (Peano).

If the function f ₁ is continuous in certain open subset, then for an arbitrary point $(t_{0},x_{0},\dot{x}_{0},\ldots,x_{0}^{(n-1)})$ belonging to this subset there is a solution to Eq.  (5.2) defined in a neighbourhood of t ₀ , which satisfies  (5.3) .

Theorem 5.2 (Cauchy–Picard).

If the function f ₁ satisfies both the conditions of Theorem  5.1 and the Lipschitz conditions with regard to the variables $x,\dot{x},\ldots,x^{(n-1)}$ , then for arbitrary initial conditions  (5.3) there is only one solution (uniqueness) to Eq.  (5.2) .

Recall that the Lipschitz constant L for the first-order case can be found from the inequality

$$\displaystyle{ \left \vert \frac{f(t,x_{1}) - f(t,x_{2})} {x_{1} - x_{2}} \right \vert = \left \vert \frac{\partial f} {\partial x}(x)\right \vert \leq L. }$$

(5.4)

In words the Lipschitz constant can be defined by an upper bound of $\,\frac{\partial f} {\partial x}$. In the nth-order case, when $f = f(t,x_{1},\ldots,x_{n})$, the Lipschitz constant L can be defined by the inequality

$$\displaystyle{ \mathrm{Max}\left \{\left \vert \frac{\partial f} {\partial x_{1}}\right \vert,\ldots,\left \vert \frac{\partial f} {\partial x_{n}}\right \vert \right \} \leq L. }$$

(5.5)

To show a uniqueness of a solution let us recall the example of the equation $\frac{\partial f} {\partial x} = 1 + x^{2}$ given in the book [191]. In this case $f(t,x_{1}) = 1 + x_{1}^{2}$ and $f(t,x_{2}) = 1 + x_{2}^{2}$. Hence $\left \vert f(t,x_{1}) - f(t,x_{2})\right \vert = \left \vert x_{1}^{2} - x_{2}^{2}\right \vert = \left \vert x_{1} - x_{2}\right \vert \left \vert x_{1} + x_{2}\right \vert \leq 2\left \vert x_{1} - x_{2}\right \vert$ in the rectangle $\left \vert t\right \vert < 1,\left \vert x\right \vert < 1$. Theorem 5.2 states that there exists only one solution passing through (0,0). In fact, this solution has the following analytical form: x(t) = tg(t). Let D be a domain composed of points, where any point corresponds to only one solution of the Cauchy problem. The function

$$\displaystyle{ x =\phi (t,C_{1},\ldots,C_{n}) }$$

(5.6)

is said to be a general solution to Eq. (5.2), if the following assumptions are satisfied:

• (i) the function ϕ has the nth-order derivative with respect to t;

• (ii) for any point defined by the series (5.3) the following equations

  $$\displaystyle\begin{array}{rcl} & x_{0} =\phi (t,C_{1},\ldots,C_{n}), & \\ & \dot{x}_{0} =\dot{\phi } (t,C_{1},\ldots,C_{n}), & \\ & \vdots & \\ & x_{0}^{(n-1)} = \phi ^{(n-1)}(t,C_{1},\ldots,C_{n}),& {}\end{array}$$

  (5.7)

  have the unique solutions with regard to the constants $C_{1}^{0},\ldots,C_{n}^{0}$;

• (iii) the function $\phi (t,C_{1}^{0},\ldots,C_{n}^{0})$ is the solution to (5.2) for arbitrary constants $C_{1}^{0},\ldots,C_{n}^{0}$, which are the solutions to (5.7).

If a general solution is given in the implicit way

$$\displaystyle{ \phi _{0}(t,x,C_{1},\ldots C_{n}) = 0, }$$

(5.8)

then Eq. (5.8) defines a general solution to Eq. (5.2).

Any arbitrary solution (5.6) for the specified values of constants $C_{1},\ldots,C_{n}$ is said to be the particular solution of Eq. (5.2). The following steps are required in order to find a particular solution knowing a general one:

• (i) the algebraic set of Eq. (5.7) should be derived from either (5.6) or (5.8), which yields the constants $C_{1},\ldots,C_{n}$;

• (ii) the found specified values $C_{1}^{0},\ldots,C_{n}^{0}$ are substituted to (5.6) or (5.8), which are now solutions to a Cauchy problem.

Note that sometimes a general solution can be represented in a parametric form

$$\displaystyle\begin{array}{rcl} t = t(p,C_{1},\ldots,C_{n}),& & \\ x = x(p,C_{1},\ldots,C_{n}),& &{}\end{array}$$

(5.9)

where p is a parameter. In the most general case governed by Eq. (5.1), the following theorem satisfies an existence and uniqueness of a Cauchy problem.

Theorem 5.3.

Assume that the function defined by  (5.1) is continuous and possesses the continuous derivatives with regard to $x,\dot{x},\ldots,x^{(n)}$ . Hence, for an arbitrary point $(t_{0},x_{0},\dot{x_{0}},\ldots,x_{0}^{(n)})$ such that

$$\displaystyle{ f(t_{0},x_{0},\dot{x}_{0},\ldots,x_{0}^{(n)}) = 0,\ \ \frac{\partial f} {\partial x^{(n)}}(t_{0},x_{0},\dot{x}_{0},\ldots,x_{0}^{(n)})\neq 0, }$$

(5.10)

there is exactly only one solution to Eq.  (5.1) defined in the neighbourhood of t ₀ and satisfying the initial conditions  (5.3) .

A general solution constitutes of a family of integral curves in the plane (t, x) with n parameters $C_{1},\ldots,C_{n}$.

5.2 Linear Homogeneous Differential Equations

Consider the following homogeneous nth-order differential equation

$$\displaystyle{ L_{n}(y) \equiv y^{(n)} + p_{ 1}(t)y^{(n-1)} + \cdots + p_{ n}(t)y = 0, }$$

(5.11)

where L _n is called the nth-order linear differential operator (here $y^{(n)} = \frac{d^{n}y} {dt^{n}}$). Recall that in general, a linear differential operator (a function) has the following properties:

• (i) L(Cy) = CL(y),

• (ii) $L(y_{1} + y_{2}) = L(y_{1}) + L(y_{2}),$

• (iii) $L(C_{1}y_{1} + \cdots + C_{m}y_{m}) = C_{1}L(y_{1}) + \cdots + C_{m}L(y_{m})$, for any y _i and C _i .

Theorem 5.4.

If $y_{1},\ldots,y_{n}$ are the solutions to homogeneous equation  (5.11) , then $y = C_{1}y_{1} + \cdots + C_{n}y_{n}$ is also a solution to L _n (y) = 0 for arbitrary constant numbers C ₁ ,…,C _n .

Remark 5.1.

The set of functions $\{y_{1}(t),\ldots,y_{n}(t)\}$ is said to be a fundamental set of solutions of L _n (y) = 0.

Remark 5.2.

The function $y(t) = C_{1}y_{1}(t) + \cdots + C_{n}y_{n}(t)$ is called the complementary function of L _n (y) = 0

Definition 5.1 (Linearly Independent Functions).

Let $u_{1}(t),\ldots,u_{m}(t)$ are the functions defined in the interval $a \leq t \leq b$. If there are numbers $\alpha _{1},\ldots,\alpha _{m}$, not all of them equal to zero, and the following equation is satisfied

$$\displaystyle{ z = \alpha _{1}u_{1} + \cdots + \alpha _{m}u_{m} \equiv 0,\qquad \ \ \ a \leq t \leq b, }$$

(5.12)

then we say that the functions $u_{1}(t),\ldots,u_{m}(t)$ are linearly dependent in the considered interval.

The vector z is called a linear combination of the members of the set $\{u_{1},\ldots,u_{m}\}$, whereas the set $\{\alpha _{1},\ldots,\alpha _{m}\}$ is called a set of coefficients of the linear combination.

Recall that:

• (i) the set of all linear combinations of members of $\{y_{1},\ldots,y_{m}\}$ is called the linear span (or simply span) of $\{y_{1},\ldots,y_{m}\}$;

• (ii) a linear space U is a set that satisfies the following properties:

  • (a) if u, v ∈ U then u + v ∈ U;

  • (b) if u ∈ U and c ∈ R, then Cu ∈ U;

• (iii) the span defined in (i) forms a linear space.

Assuming that Eq. (5.12) is satisfied if and only if $\alpha _{1} = \cdots =\alpha _{m} \equiv \ 0$, then the functions $u_{1},\ldots,u_{m}$ are called linearly independent. A linearly independent spanning set for a linear space $U =\{ u_{1},\ldots,u_{m}\}$ is called a basis for U. A dimension of U is defined by the number of vectors in a basis of vector space U. A necessary condition for linear dependence of the functions $u_{1},\ldots,u_{n}$ can be obtained in the following way. Let the given functions $u_{1},\ldots,u_{n}$ to be are linearly dependent. It means that they satisfy (5.12). Differentiating n − 1 times Eq. (5.12) one gets

$$\displaystyle{ \alpha _{1}u_{1}^{(k)} + \cdots + \alpha _{ n}u_{n}^{(k)} = 0,\quad k = 1,\ldots,n - 1. }$$

(5.13)

Of course, we have assumed that the functions $\{u_{1},\ldots,u_{n}\}$ have n − 1 continuous derivatives on some interval (they are C ⁿ⁻¹ smooth). It is known from algebra that the system of Eqs. (5.12) and (5.13) has non-trivial solution of $\alpha _{1},\ldots,\alpha _{n}$, when the following determinant function (called Wronskian) is equal to zero:

$$\displaystyle{ W(u_{1},\ldots,u_{n}) = \left \vert \begin{array}{*{10}c} u_{1} & \cdots & u_{n} \\ u_{1}^{{\prime}} &\cdots & u_{n}^{{\prime}}\\ \vdots \\ u_{1}^{(n-1)} & \cdots &u_{n}^{(n-1)}\\ \end{array} \right \vert = 0,\quad a \leq t \leq b. }$$

(5.14)

Theorem 5.5.

If the functions $y_{1},\ldots,y_{n}$ are the solutions of L _n (y) = 0 on a ≤ t ≤ b, then $W(y_{1}(t),\ldots,y_{n}(t))$ is either zero for every t from [a,b] or is never zero for any t ∈ [a,b].

Proof.

We take n = 2 and follow the proof given in the book [191]. Here we omit a generalization for any n. Consider the following linear operator with variable coefficients

$$\displaystyle{L_{2}(y) \equiv y^{{\prime\prime}} + p_{ 1}(t)y^{{\prime}} + p_{ 2}(t)y = 0.}$$

The Wronskian

$$\displaystyle{W(y_{1},y_{2}) = \left \vert \begin{array}{*{10}c} y_{1} & y_{2} \\ y_{1}^{{\prime}}&y_{2}^{{\prime}}\\ \end{array} \right \vert = y_{1}y_{2}^{{\prime}}-y_{ 2}y_{1}^{{\prime}}.}$$

On the other hand

$$\displaystyle\begin{array}{rcl} W^{{\prime}}& =& (y_{ 1}y_{2}^{{\prime}}- y_{ 2}y_{1}^{{\prime}})^{{\prime}} = y_{ 1}y^{{\prime}}_{ 2}{}^{{\prime}}-y^{{\prime}}_{ 1}{}^{{\prime}}y_{ 2} = y_{1}(-p_{1}y_{2}^{{\prime}}- p_{ 2}y_{2}) - y_{2}(-p_{1}y_{1}^{{\prime}}- p_{ 2}y_{1}) {}\\ & =& -p_{1}(y_{1}y_{2}^{{\prime}}- y_{ 2}y_{1}^{{\prime}}) = -p_{ 1}W, {}\\ \end{array}$$

and hence

$$\displaystyle{\ln W = -\int p_{1}dt +\ln C.}$$

Finally, one gets

$$\displaystyle{W(y_{1},y_{2}) = Ce^{-\int p_{1}(t)dt},}$$

which for C = 0 is identically zero, or for C ≠ 0 is never zero. □ 

Theorem 5.6.

If $y_{1},\ldots,y_{n}$ are solutions of the equation L _n (y) = 0 and $\left.W(y_{1},\ldots,y_{n})\right \vert _{t=t_{0}} = 0$ , then $W(y_{1},\ldots,y_{n}) \equiv 0$ for a ≤ t ≤ b, and $y_{1},\ldots,y_{n}$ are linearly dependent solutions in the interval t ∈ [a,b].

Theorem 5.7.

A necessary and sufficient condition for solutions $z_{1},\ldots,z_{n}$ of L _n (z) = 0 to be linearly independent in an arbitrary point is $\left.W(z_{1},\ldots,z_{n})\right \vert _{t=t_{0}}\neq 0$ . If W(t ₀ ) ≠ 0, then also W(t) ≠ 0 for a ≤ t ≤ b, and if W(t ₀ ) = 0, then also W(t) = 0.

Theorem 5.8.

If $y_{1},\ldots,y_{n}$ are linearly independent solutions of the equation L _n (y) = 0, then $y = C_{1}y_{1} + \cdots + C_{n}y_{n}$ is a general solution for a ≤ t ≤ b, $-\infty < y_{k} < +\infty$, $k = 0,1,\ldots,n - 1$ .

In the case of a homogeneous linear nth-order equation L _n (y) = 0, there are n linearly independent solutions $y_{1},\ldots,y_{n}$ forming the fundamental set of solutions (proof is omitted here). It is interesting to note that the Wronskian can be used to define a differential equation, when a fundamental set is known.

Recall that a set of all solutions of the linear equation L(y) = 0 is called a null space or kernel of the operator L.

Theorem 5.9.

A homogeneous linear differential equation L _n (y) = 0 has n linearly independent solutions that form a basis for the set of all solutions. The dimension of the kernel of an n th-order linear differential operator is n.

Example 5.1.

Define a linear differential equation possessing the fundamental solution: y ₁(t) = sint, y ₂(t) = e ^t.

We define the Wronskian

$$\displaystyle\begin{array}{rcl} W(\sin t,e^{t},y)& =& \left \vert \begin{array}{*{10}c} \sin t &e^{t}& y \\ \cos t &e^{t}& y^{{\prime}} \\ -\sin t&e^{t}&y^{{\prime\prime}}\\ \end{array} \right \vert e^{t}\left \vert \begin{array}{*{10}c} \sin t & 1 & y \\ \cos t & 1 &y^{{\prime}} \\ -\sin t&1\&y^{{\prime\prime}}\\ \end{array} \right \vert {}\\ & =& e^{t}\left (\sin t\left \vert \begin{array}{*{10}c} 1& y^{{\prime}} \\ 1&y^{{\prime\prime}}\\ \end{array} \right \vert - 1\left \vert \begin{array}{*{10}c} \cos t & y^{{\prime}} \\ -\sin t&y^{{\prime\prime}}\\ \end{array} \right \vert + y\left \vert \begin{array}{*{10}c} \cos t &1\\ -\sin t &1\\ \end{array} \right \vert \right ) {}\\ & =& e^{t}[\sin t(y^{{\prime\prime}}- y^{{\prime}}) - (y^{{\prime\prime}}\cos t + y^{{\prime}}\sin t) + y(\cos t +\sin t)] {}\\ & =& e^{t}[(\sin t -\cos t)y^{{\prime\prime}}- 2\sin t + y^{{\prime}} + (\sin t +\cos t)y] = 0, {}\\ \end{array}$$

and the being sought differential equation is defined immediately as

$$\displaystyle{L_{2}(y) = (\sin t -\cos t)y^{{\prime\prime}}- 2\sin ty^{{\prime}} + (\sin t +\cos t)y = 0.}$$

In order to verify the obtained results we check:

$$\displaystyle\begin{array}{rcl} & L_{2}(\sin t) = -(\sin t -\cos t)\sin t - 2\sin t\cos t + (\sin t +\cos t)\sin t = 0,& {}\\ & L_{2}(e^{t}) = e^{t}L_{2}(\sin t) = 0. & {}\\ \end{array}$$

 □ 

5.3 Differential Equations with Constant Coefficients

Since characteristic equation corresponding to an nth-order homogeneous differential equation cannot be solved using radicals already for n ≥ 5, very often numerical or approximate analytical methods are used. The most important observation is that the nth-order differential equation is reduced to the nth-order polynomial (algebraic) equation. The solutions differ from each other qualitatively for distinct and repeated (multiple) roots of a characteristic equation, which we are going to discuss.

Theorem 5.10.

Given the n th-order linear differential equation $L_{n}(y) = y^{(n)} + p_{1}y^{(n-1)} + \cdots + p_{n-1}y^{{\prime}} + p_{n}y = 0.$ Let the characteristic polynomial corresponding to L _n (y) = 0 possess:

• (i) n distinct real roots r _i , i = 1,…,n. Then the n functions $y_{i}(t) = e^{r_{i}t}$ form a basis for the kernel of L _n (y) = 0. A general solution has the form

  $$\displaystyle{y(t) =\sum \limits _{ i=1}^{n}C_{ i}e^{r_{i} t};}$$

• (ii) Single root repeated n times. Then the n functions $y_{1}(t) = e^{rt},y_{2}(t) = te^{rt},\ldots,y_{n}(t) = t^{n-1}e^{rt}$ form a basis for the kernel of L _n (y) = 0. A general solution can be written as

  $$\displaystyle{y(t) = (C_{1} + c_{2}t + \cdots + C_{n}t^{n-1})e^{rt};}$$

• (iii) Single root repeated m times and there are n − m distinct roots $r_{m+1},r_{m+2},\ldots,r_{n}.$ Then the n functions

  $$\displaystyle{\begin{array}{*{10}c} y_{1}(t) = e^{rt}, \\ y_{2}(t) = te^{rt},\\ \vdots \\ y_{m}(t) = t^{m-1}e^{rt}, \\ y_{m+1}(t) = e^{r_{m+1}t},\\ \vdots \\ y_{n}(t) = e^{r_{n}t} \\ \end{array} }$$

  form the kernel of L _n (y) = 0. A general solution has the form

  $$\displaystyle{y(t) = (C_{1} + C_{2} + \cdots + C_{m}t^{m-1})e^{rt} + C_{ m+1}e^{r_{m+1} t} + \cdots + C_{ n}e^{r_{n} t}.}$$

Example 5.2.

Consider the following third-order differential equation

$$\displaystyle{L_{3}(y) = y^{{\prime\prime\prime}}- 8y^{{\prime\prime}} + 19y^{{\prime}}- 12y = 0.}$$

Introducing y = e ^rt one obtains the characteristic polynomial $r^{3} - 8r^{2} + 19r - 12 = 0$, with three distinct roots: $r_{1} = 1,r_{2} = 3,r_{3} = 4$. The linearly independent solutions are: $y_{1}(t) = e^{t},y_{2}(t) = e^{3t},y_{3}(t) = e^{4t}$, and every solution is written as $y(t) = C_{1}e^{t} + C_{2}e^{3t} + C_{3}e^{4t}$. □ 

Example 5.3.

Consider the fifth-order differential equation $L_{5}(y) = y^{(5)} - 12y^{(4)} + 56y^{{\prime\prime\prime}}- 126y^{{\prime\prime}} + 135y^{{\prime}}- 54 = 0.$

The characteristic equation corresponding to L ₅(y) has the following roots: r ₁ = 1, r ₂ = 2 and $r_{3} = r_{4} = r_{5} = 3$ (multiplicity 3). The following set forms a basis $\{e^{t},\ e^{2t},\ e^{3t},\ te^{3t},\ t^{2}e^{3t}\}$. Every solution in the kernel has the form

$$\displaystyle{y(t) = C_{1}e^{t} + C_{ 2}e^{2t} + C_{ 3}e^{3t} + C_{ 4}te^{3t} + C_{ 5}t^{2}e^{3t}.}$$

In order to investigate their linear dependence one calculates

$$\displaystyle\begin{array}{rcl} & W(e^{t},e^{2t},e^{3t},te^{3t},t^{2}e^{3t}) & {}\\ & = \left \vert \begin{array}{*{10}c} e^{t}& e^{2t} & e^{3t} & te^{3t} & t^{2}e^{3t} \\ e^{t}& 2e^{2t} & 3e^{3t} & e^{3t} + 3te^{3t} & 2te^{3t} + 3t^{2}e^{3t} \\ e^{t}& 4e^{2t} & 9e^{3t} & 6e^{3t} + 9te^{3t} & 2e^{3t} + 12te^{3t} + 9t^{2}e^{3t} \\ e^{t}& 8e^{2t} &27e^{3t}& 27e^{3t} + 27te^{3t} & 16e^{3t} + 54te^{3t} + 27t^{2}e^{3t} \\ e^{t}&16e^{2t}&81e^{3t}&108e^{3t} + 81te^{3t}&108e^{3t} + 216te^{3t} + 81t^{2}e^{3t}\\ \end{array} \right \vert & {}\\ & = e^{t}e^{2t}e^{3t}e^{3t}e^{3t}\left \vert \begin{array}{*{10}c} 1& 1 & 1 & t & t^{2} \\ 1& 2 & 3 & 1 + 3t & 2t + 3t^{2} \\ 1& 4 & 9 & 6 + 9t & 2 + 12t + 9t^{2} \\ 1& 8 &27& 27 + 27t & 16 + 54t + 27t^{2} \\ 1&16&81&108 + 81t&108 + 216t + 81t^{2}\\ \end{array} \right \vert \neq 0.& {}\\ & & {}\\ \end{array}$$

Theorem 5.11.

Given the n th-order linear differential equation $L_{n}(y) = y^{(n)} + p_{1}y^{(n-1)} + \cdots + p_{n-1}y^{{\prime}} + p_{n}y = 0$ . Let the characteristic equation corresponding to L _n (y) = 0 to have:

• (i) A pair of complex roots $r = a \pm bi,\quad i^{2} = -1$ . Then the functions $y_{1}(t) = e^{at}\cos bt$ and $y_{2}(t) = e^{at}\sin bt$ are linearly independent solution of L _n (y) = 0.

• (ii) A pair of complex conjugate roots repeated m times. Then the 2m functions

  $$\displaystyle\begin{array}{rcl} & \quad y_{1}(t) = e^{at}\cos bt,\quad y_{2}(t) = te^{at}\cos bt\quad,\ldots,\quad y_{m}(t) = t^{m-1}e^{at}\cos bt, & {}\\ & y_{m+1}(t)=e^{at}\sin bt,\quad y_{m+2}(t)=te^{at}\sin bt\quad,\ldots,\quad y_{2m}(t) = t^{m-1}e^{at}\sin bt& {}\\ \end{array}$$

  are linearly independent solutions of L _n (y) = 0. In addition, every function $y(t) = (C_{1} + C_{2}t + \cdots + C_{m}t^{m-1})e^{at}\cos bt + (C_{m+1} + C_{m+2}t + \cdots + C_{2m}t^{m-1})e^{at}\sin bt$ is a solution to the studied equation.

In order to get a better imagination of understanding of repeated roots of a characteristic equation we review some properties of polynomials.

• (i) A polynomial w(t) has a root r if $w(t) = (t - r)w_{0}(t)$.

• (ii) A polynomial w(t) has a double root if $w(t) = (t - r)^{2}w_{1}(t)$. In this case w(t) and $w^{{\prime}}(t) = 2(t - r)w_{1}(t)$ have a common root r.

• (iii) In general, a polynomial w(t) has a root r of multiplicity k if $w(t) = (t - r)^{k}w_{k-1}(t)$. In this case, $w(t),w^{{\prime}}(t),\ldots,w^{k-1}(t)$ have a common root r.

Now we focus our attention on a second polynomial corresponding to differential operator $L_{2}(y) = y^{{\prime\prime}}- 2ry^{{\prime}} + r^{2}y$. The corresponding quadratic equation is $w(t) = t^{2} - 2rt + r^{2} = (t - r)^{2}$. According to Theorem 5.10 we have the solutions: $y_{1}(t) = e^{rt},y_{2}(t) = te^{rt}$. It is true, because $L_{2}(y_{1}) = e^{rt}(r^{2} - 2r^{2} + r^{2}) = 0$, and $L_{2}(te^{rt}) = (te^{rt})^{{\prime\prime}}-2r(te^{rt})^{{\prime}}+r^{2}te^{rt} = (e^{rt}+rte^{rt})^{{\prime}}-2r(e^{rt}+rte^{rt})+r^{2}te^{rt} = 2re^{rt}+r^{2}te^{rt}-2re^{rt}-2r^{2}te^{rt}+r^{2}te^{rt} = 0$. The above example leads to more general observation. Namely, have got $L_{2}(e^{rt}) = e^{rt}w(r)$. Observe that $\frac{\partial } {\partial r}L_{2}(e^{rt}) = \frac{\partial } {\partial r}[e^{rt}w(r)] = te^{rt}w(r) + w^{{\prime}}(r)e^{rt}.$ But recall that r is the double root of the characteristic equation w(r) = 0 and hence $w(r) = w^{{\prime}}(r) = 0$, i.e. $\frac{\partial } {\partial r}L_{2}(e^{rt}) = 0.$ On the other hand $\frac{\partial } {\partial r}L_{2}(e^{rt}) = L_{ 2}( \frac{\partial } {\partial r}e^{rt}) = L_{ 2}(te^{rt}) = 0,$ which shows that te ^rt is a solution of the differential operator L ₂. Finally, let us check a linear dependence of both solutions. The Wronskian determinant gives

$$\displaystyle{w(e^{rt},te^{rt}) = \left \vert \begin{array}{*{10}c} e^{rt}\qquad \quad te^{rt} \\ e^{rt}\quad e^{rt} + te^{rt}\\ \end{array} \right \vert = e^{2rt}\neq 0,}$$

which means that they are independent. This observation can be generalized by the following theorem.

Theorem 5.12.

If r ₀ is a root of m-multiplicity of the characteristic equation

$$\displaystyle{w(r) = r^{n} + a_{ 1}r^{n-1} + a_{ 2}r^{n-2} + \cdots + a_{ n-1}r + a_{n} = 0,}$$

where 1 ≤ m ≤ n, then the functions $e^{r_{0}t},te^{r_{0}t},\ldots,r^{m-1}e^{r_{0}t}$ are the linearly independent solutions of the characteristic equation w(r).

Proof.

If r ₀ is the root of w(r) of multiplicity m, then

$$\displaystyle{w(r_{0}) = \frac{dw} {dr} (r_{0}) = \frac{d^{2}w} {dr^{2}} (r_{0}) = \cdots = \frac{d^{k-1}w(r_{0})} {dr^{k-1}} = 0.}$$

Consider now the differential operator

$$\displaystyle{L(y) \equiv y^{(n)} + a_{ 1}y^{(n-1)} + \cdots + a_{ n}y.}$$

It is easy to check that $L(e^{rt}) = w(r)e^{rt}$. Consider now $L(t^{i}e^{rt}) = L\left (\frac{\partial ^{i}e^{rt}} {\partial r^{i}} \right ) = \frac{\partial ^{i}} {\partial r^{i}}L(e^{rt}) = \frac{\partial ^{i}} {\partial r^{i}}\left (w(r)e^{rt}\right ).$ We have

$$\displaystyle{\begin{array}{*{10}c} i = 0:& w(r)e^{rt} = 0; \\ i = 1:& (w^{{\prime}} + tw)e^{rt} = 0; \\ i = 2:&(w^{{\prime\prime}} + 2tw^{{\prime}} + w(1 + t^{2}))e^{rt} = 0;\\ \vdots\\ \end{array} }$$

Since for $i = k - 1$ we obtain $w^{k-1} = \cdots = w^{{\prime\prime}} = w^{{\prime}} = w = 0$, hence we have also $L(t^{i}e^{r_{0}t}) = 0$. It means that $t^{i}e^{r_{0}t}$ are solutions to L(y). In addition, they are linearly independent, because the polynomials of different orders are linearly independent. □ 

In the case of complex conjugate roots (Theorem 5.12) we recall some fundamental properties known from algebra. According to the Euler’s formula we have

$$\displaystyle{e^{i\beta } =\cos \beta +i\sin \beta,}$$

where: Re(e ^iβ) = cosβ, $\mathop{\mathrm{Im}}\nolimits (e^{i\beta }) =\sin \beta$ and $e^{a+ib} = e^{a}(\cos b + i\sin b).$

Now let the characteristic equation representing the differential equation L(y) = 0 to have conjugate complex roots a ± ib, and we obtain the following solutions to the differential equation: $e^{(a+ib)t} = e^{at}(\cos bt + i\sin bt)$ and $e^{(a-ib)t} = e^{at}(\cos bt - i\sin bt)$. Observe that instead of taking this solution one can take $y_{1} = e^{at}\cos bt$ and $y_{2} = e^{at}\sin bt$. Both of them are real-valued independent solutions (check), and they span the same real-valued solutions as the complex-valued solutions e ^(a±ib)t. This observation can be generalized by the following theorem.

Theorem 5.13.

If u and v are in the domain composed of real elements of a real-valued function L, and we take $y = u + iv$ , then $L(y) = L(u) + iL(v)$ . In addition, if L(y) = 0, then $L(u) = L(v) = 0$ .

Remark 5.3.

Theorem 5.12 shows that the domain of L can be extended to include complex-valued component.

Our considerations can be briefly outlined in the following way [191]. The solutions to homogeneous linear nth-order ordinary differential equation (5.11) are sought in forms of exponential functions

$$\displaystyle{ y = e^{rt}, }$$

(5.15)

where r is constant. We have

$$\displaystyle{ y^{(k)} = \frac{\partial ^{k}y} {\partial t^{k}} = r^{k}e^{rt},\quad k = 1,\ldots,n. }$$

(5.16)

Substituting the assumed solution (5.15) to (5.16) we get

$$\displaystyle{ L(e^{rt}) = e^{rt}P_{ n}(r) = 0, }$$

(5.17)

where the polynomial

$$\displaystyle{ P_{n}(r) = r^{n} + p_{ 1}r^{n-1} + \cdots + p_{ n-1}r + p_{n} = 0 }$$

(5.18)

is called the characteristic equation (or the characteristic polynomial). The roots of the characteristic equation (5.18) are called the characteristic roots. If the roots $r_{1},\ldots,r_{n}$ are real and different, then we have n independent solutions of the form

$$\displaystyle{ y_{1} = e^{r_{1}t},\ldots,y_{ n} = e^{r_{n}t}. }$$

(5.19)

The Wronsky determinant has the form

$$\displaystyle\begin{array}{rcl} w(e^{r_{1} t},\ldots,e^{r_{n} t})& =& \left \vert \begin{array}{*{10}c} e^{r_{1}t} &\ldots & e^{r_{n}t} \\ r_{1}e^{r_{1}t} &\ldots & r_{n}e^{r_{n}t}\\ \vdots &\vdots & \vdots \\ r_{1}^{n-1}e^{r_{1}t}&\ldots &r_{n}^{n-1}e^{r_{n}t} \end{array} \right \vert \\ & =& e^{(r_{1}+r_{2}+\cdots +r_{n-1})t}\left \vert \begin{array}{*{10}c} 1 & 1 &\ldots & 1\\ r_{1 } & r_{2 } & \ldots & r_{n}\\ \vdots & \vdots &\vdots & \vdots \\ r_{1}^{n-1} & r_{2}^{n-1} & \ldots & r_{n}^{n-1} \end{array} \right \vert \\ & =& e^{(r_{1}+r_{2}+\cdots +r_{n-1})t}(r_{ n} - r_{1})(r_{n} - r_{2})\ldots (r_{n} - r_{n-1})(r_{n-2} - r_{1})\ldots \\ & & \ldots (r_{n-1}-r_{n-2})\ldots (r_{2}-r_{1})\neq 0. {}\end{array}$$

(5.20)

Observe that w(t) → 0, if t → +∞, when $r_{k} < 0,k = 1,\ldots,n.$ A general (or every) solution to (5.11) has the form

$$\displaystyle{ y = C_{1}e^{r_{1}t} + \cdots + C_{ n}e^{r_{n}t}. }$$

(5.21)

If r _k  < 0, k = 1, …, n, then every solution y governed by (5.21) y → 0 for t → +∞.

Consider the case, when $r_{1},\ldots,r_{n}$ are different, but some of them are complex conjugate. Let $r_{1} = a + ib$. Since the coefficients of the differential equation (5.11) are real, then also $r_{2} = \bar{r}_{1} = a - ib$ exists. A general solution has the form $y = u(t) + iv(t)$, where u(t) and v(t) are real-valued functions. However, if $y(t_{0}) = u(t_{0}) + iv(t_{0}) = y_{0},\ y^{(k)}(t_{0}) = u^{(k)}(t_{0}) + iv^{(k)}(t_{0}) = y_{0}^{(k)},\ k = 1,\ldots,n - 1$, where y ₀ and $y_{0}^{(k)}$ are real-valued, then $v(t_{0}) = v^{{\prime}}(t_{0}) = \cdots = v^{(n-1)}(t_{0}) = 0$, and hence v(t) ≡ 0.

Given the initial conditions one finds the being sought constants $C_{1},\ldots,C_{n}$ from the algebraic equations

$$\displaystyle\begin{array}{rcl} y_{0}& =& C_{1}e^{r_{1} t_{0} } + \cdots + C_{n}e^{r_{1} t_{0} }, \\ y_{0}^{(k)}& =& C_{ 1}r_{1}^{k}e^{r_{1} t_{0} } + \cdots + C_{n}r_{n}^{k}e^{r_{n} t_{0} },\quad k = 1,\ldots,n - 1.{}\end{array}$$

(5.22)

It can be shown that the solution (5.21), after substituting the constant $C_{i}i = 1,\ldots,n$ from (5.22), is real-valued. Hence, since the solution is real-valued then instead of complex-valued form of the solution we can use the real-valued form of the solution. For our case $r_{1} = a + ib$, $r_{2} = a - ib$ we get

$$\displaystyle\begin{array}{rcl} & & C_{1}e^{r_{1} t} + C_{ 2}e^{r_{2} t} = C_{ 1}e^{at}(\cos bt + i\sin bt) + C_{ 2}e^{at}(\cos bt - i\sin bt) {}\\ & & \quad = (C_{1} + C_{2})e^{at}\cos bt + i(C_{ 1} - C_{2})e^{at}\sin bt = C_{ 10}e^{at}\cos bt + C_{ 20}\sin bt, {}\\ \end{array}$$

where: $C_{10} = C_{1} + C_{2},C_{20} = i(C_{1} - C_{2})$. It is easy to check that $C_{1,2} = C_{10} \pm C_{20}/$ i, i.e. they are complex conjugate, but C _i0, i = 1, 2 are real-valued. In other words

$$\displaystyle{ y = C_{10}e^{at}\cos bt + C_{ 20}e^{at}\sin bt + C_{ 3}e^{r_{3}t} + \cdots + C_{ n}e^{r_{n}t}. }$$

(5.23)

For real-value initial conditions $y_{0},\ y_{0}^{{\prime}},\ldots,\ y^{(n-2)}$ the solution (5.23) is real-valued and also C ₁₀ and C ₂₀ should be real-valued. Since $e^{r_{1}t} = e^{at}(\cos bt + i\sin bt)$ is the solution, then also $y_{1} = e^{at}\cos bt$ and $y_{2} = e^{at}\sin bt$ are the solutions generated by the roots $r_{1} = a + ib,r_{2} = a - bt$. Recall now the discussed case of the root $r_{1} = \cdots = r_{m}$ with mth-multiplicity. The kth-order derivative of (5.11) yields

$$\displaystyle{ \frac{d^{k}L(e^{rt})} {dr^{k}} = L\left [\frac{d^{k}e^{rt}} {dr^{k}} \right ] = L(t^{k}e^{rt}), }$$

(5.24)

and

$$\displaystyle{ L(t^{k}e^{rt}) = (t^{k}e^{rt})^{(n)} + p_{ 1}(t^{k}e^{rt})^{n-1} + \cdots + p_{ n-1}(t^{k}e^{rt})^{{\prime}} + p_{ n}(t^{k}e^{rt}). }$$

(5.25)

Owing to the Leibniz formula applied to dot product of two functions we have

$$\displaystyle{ (uv)^{(n)} = uv^{(n)} + \left (\begin{array}{*{10}c} n\\ 1\\ \end{array} \right )u^{{\prime}}v^{(n-1)} + \left (\begin{array}{*{10}c} n\\ 2\\ \end{array} \right )u^{{\prime\prime}}v^{(n-2)}+\ldots \ldots + \left (\begin{array}{*{10}c} n\\ n - 1\\ \end{array} \right )u^{(n-1)}v^{{\prime}} + u^{(n)}v =\sum \limits _{ k=0}^{n}u^{(k)}v^{(n-k)}, }$$

(5.26)

where: u ⁽⁰⁾ = u, v ⁽⁰⁾ = v. Applying Leibniz formula (5.26) to (5.25) we obtain

$$\displaystyle\begin{array}{rcl} & & L(t^{k}e^{rt}) = t^{k}e^{rt}P_{ n}(r) + C_{1}^{(k)}t^{k-1}e^{rt}P^{{\prime}}_{ n}(r) \\ & & \,\qquad \qquad + C_{2}^{(k)}t^{k-2}e^{rt}P^{{\prime\prime}}_{ n}(r) + \cdots + e^{rt}P_{ n}^{(k)}(r),{}\end{array}$$

(5.27)

where the numbers $C_{l}^{(k)}$ depend on k and l. Since $r_{1} = \cdots = r_{m}$, hence the characteristic equation has the following property

$$\displaystyle{ P_{n}(r_{1}) = P_{n}^{{\prime}}(r_{ 1}) = \cdots = P_{n}^{m-1}(r_{ 1}) = 0. }$$

(5.28)

It means that $L(t^{k}e^{rt}) = 0$ for $k = 1,2,\ldots,m - 1$ and r = r ₁, where r ₁ is the root of mth multiplicity. In other words, m times repeated root generates m solutions of the form $e^{r_{1}t},te^{r_{1}t},\ldots,t^{m-1}e^{r_{1}t}.$ If $r_{1},\ldots,r_{l}$ are different roots of the corresponding multiplicity $m_{1},\ldots,m_{l}$ and there are no other roots (i.e. $m_{1} + \cdots + m_{l} = n$), then every solution to homogeneous linear differential equation (5.11) has the form

$$\displaystyle{ y = P_{m_{1}-1}(t)e^{r_{1} t} + P_{ m_{2}-1}(t)e^{r_{2} t} + \cdots + P_{ m_{l}-1}(t)e^{r_{l} t}, }$$

(5.29)

where: $P_{m_{1}-1}(t),P_{m_{2}-1}(t),\ldots,P_{m_{l}-1}(t)$ are the polynomials of t of the corresponding orders: $m_{1} - 1,m_{2} - 1,\ldots,m_{l} - 1$. It can be checked that a set $\{e^{r_{1}t},re^{r_{1}t},\ldots,t^{m_{1}-1}e^{r_{1}t},\ldots,e^{r_{l}t},te^{r_{l}t},\ldots,t^{m_{l}-1}e^{r_{l}t}\}$ has members which are linearly independent.

Recall now our considerations related to repeated complex roots. If among the roots $r_{1},\ldots,r_{l}$ a complex one appears, say $r_{1} = a + ib$, then also exists $r_{2} = a - ib$ with the same multiplicity. The used real initial conditions $y_{0},y_{0}^{{\prime}},\ldots,y_{0}^{(n-1)}$ allow to find a set of constants to define a real solution.

Assume that $r = a + ib$ and $\bar{r} = a - ib$ are m times repeated roots. Every solution includes the functions

$$\displaystyle\begin{array}{rcl} & e^{rt},te^{rt},\ldots,t^{m-1}e^{rt},& \\ & e^{\bar{r}t},te^{\bar{r}t},\ldots,t^{m-1}e^{\bar{r}t}.&{}\end{array}$$

(5.30)

Since $t^{k}e^{rt} = t^{k}e^{at}e^{ibt} = t^{k}e^{at}(\cos bt + i\sin bt)$ is the solution, then its real and imaginary part are also solutions. The same consideration holds for $\bar{r}$. In the case when a root is purely imaginary with mth multiplicity (a = 0), then every solution is composed of the functions $t^{k}\cos bt,t^{k}\sin bt,k = 0,1,\ldots,m.$

Let us sum up our brief conclusions related to a repeated root. Any real-valued root r with mth multiplicity generates m solutions, whereas a complex-valued root r with mth multiplicity generates 2m solutions. Every solution corresponding to m-times repeated root is represented by

$$\displaystyle{ y =\sum \limits _{ k=0}^{m-1}C_{ k}t^{k}e^{rt}. }$$

(5.31)

Every solution corresponding to complex root with mth multiplicity has the form

$$\displaystyle{ y =\sum \limits _{ k=0}^{m-1}\left (A_{ k}t^{k}e^{at}\cos bt + B_{ k}t^{k}e^{at}\sin bt\right ). }$$

(5.32)

Finally, every solution corresponding to repeated imaginary roots r = ib, $\bar{r} = -ib$ has the form

$$\displaystyle{ y =\sum \limits _{ k=0}^{m-1}\left (A_{ k}t^{k}\cos bt + B_{ k}t^{k}\sin bt\right ). }$$

(5.33)

In both latter cases, A _k and B _k are defined by the initial conditions.

Example 5.4.

Using formula (5.25) and (5.26) for k = 2 find a corresponding number $C_{l}^{k}$ when

$$\displaystyle{L(y) = y^{{\prime\prime\prime}} + p_{ 1}y^{{\prime\prime}} + p_{ 2}y^{{\prime}} + p_{ 3}y.}$$

Formula (5.26) gives

$$\displaystyle\begin{array}{rcl} & (uv)^{{\prime\prime\prime}} = uv^{{\prime\prime\prime}} + \left (\begin{array}{*{10}c} 3\\ 1\\ \end{array} \right )u^{{\prime}}v^{{\prime\prime}} + \left (\begin{array}{*{10}c} 3\\ 2\\ \end{array} \right )u^{{\prime\prime}}v^{{\prime}} + u^{{\prime\prime\prime}}v,& {}\\ & (uv)^{{\prime\prime}} = uv^{{\prime\prime}} + \left (\begin{array}{*{10}c} 2\\ 1\\ \end{array} \right )u^{{\prime}}v^{{\prime}} + u^{{\prime\prime}}v^{{\prime}}, & {}\\ & (uv)^{{\prime}} = u^{{\prime}}v + v^{{\prime}}y, & {}\\ & & {}\\ \end{array}$$

whereas from (5.25) we get

$$\displaystyle\begin{array}{rcl} L(t^{k}e^{rt})& =& (t^{k}e^{rt})^{{\prime\prime}} + p_{ 1}(t^{k}e^{rt})^{{\prime\prime}} + p_{ 2}(t^{k}e^{rt})^{{\prime}} + p_{ 3}(t^{k}e^{rt}) {}\\ & =& (t^{2}r^{3}+6tr^{2}+6r)e^{rt}+p_{ 1}(t^{2}r^{2}+4tr+2)e^{rt}+p_{ 2}(t^{2}r+2t)e^{rt}+p_{ 3}t^{2}e^{rt} {}\\ & =& e^{rt}[t^{2}(r^{3} + p_{ 1}r^{2} + p_{ 2}r + p_{3}) + t(6r^{2} + 4rp_{ 1} + 2p_{2}) + 6r + 2p_{1}]. {}\\ \end{array}$$

Since

$$\displaystyle\begin{array}{rcl} P_{3}(r)& =& r^{3} + p_{ 1}r^{2} + p_{ 2}r + p_{3}, {}\\ P^{{\prime}}_{ 3}(r)& =& 3r^{2} + 2p_{ 1}r + p_{2}, {}\\ P^{{\prime\prime}}_{ 3}(r)& =& 6r + 2p_{1}, {}\\ \end{array}$$

therefore

$$\displaystyle{L(t^{2}e^{rt}) = t^{2}e^{rt}P_{ 3}(r) + C_{1}^{2}P_{ 3}^{{\prime}}(r)e^{rt} + e^{rt}P_{ 3}^{{\prime\prime}}(r),}$$

and we have $C_{1}^{2} = 2$.

5.4 Linear Non-homogeneous Differential Equations with Constant Coefficients

In the previous section we have considered linear homogeneous operator defined by Eq. (5.11). Here we are going to consider non-homogeneous problems.

Theorem 5.14.

If the fundamental solutions $y_{1}(t),\ldots,y_{n}(t)$ to Eq.  (5.11) are known, then a general solution to the following non-homogeneous equation

$$\displaystyle{ L_{n}(y) = y^{(n)} + p_{ 1}(t)y^{(n-1)} + \cdots + p_{ n}(t)y = f(t) }$$

(5.34)

has the form

$$\displaystyle{ y = y_{p} + C_{1}y_{1} + C_{2}y_{2} + \cdots + C_{n}y_{n}, }$$

(5.35)

where $C_{1},C_{2},\ldots,C_{n}$ are arbitrary numbers, and y _p is an arbitrary particular solution of  (5.34) .

There are two general methods to solve a non-homogeneous problem, i.e. the method of undetermined coefficients and the method of variations of parameters. The method of undetermined coefficients relies on observation that a being sought solution is somehow similar to the right-hand side function f(t).

Example 5.5.

Find a general solution to the non-homogeneous differential equation

$$\displaystyle{L(y) = y^{{\prime\prime}}- 3y^{{\prime}} + 2y = 3t^{2} - 2t + 4.}$$

First we consider the corresponding homogeneous equation $L(y) = y^{{\prime\prime}}- 3y^{{\prime}} + 2y = 0$. The corresponding characteristic equation is $r^{2} - 3r + 2 = 0$ with the roots r ₁ = 1 and r ₂ = 2, and hence the general solution of the homogeneous equation $y = C_{1}e^{t} + C_{2}e^{2t}$. Let us look for a particular solution of the form $y_{p} = at^{2} + bt + c.$ The simple calculus yields

$$\displaystyle\begin{array}{rcl} & L(y_{p}) = 2at - 3(2at + b) + 2(at^{2} + bt + c)& {}\\ & = 2at^{2} + t(2b - 6a) + 2c + 2a - 3b. & {}\\ \end{array}$$

Equating the coefficients standing by the same powers of t we obtain:

$$\displaystyle{\begin{array}{*{10}c} t^{0}:&2c + 2a - 3b = 4, \\ t^{1}:& 2b - 6a = -2, \\ t^{2}:& 2a = 3.\\ \end{array} }$$

The above linear algebraic equations give: $a = 1.5,\ b = 3.5,\ c = 5.75$. The general solution of the non-homogeneous problem is $y = C_{1}e^{t} + C_{2}e^{2t} + 1.5t^{2} + 3.5t + 5.75$. □ 

Example 5.6.

Find a general solution of the non-homogeneous equation

$$\displaystyle{L(y) = y^{{\prime\prime}} + 2hy^{{\prime}} +\alpha _{ 0}^{2}y = q\cos \omega t.}$$

This equation governs oscillation of a one-degree-of-freedom mechanical system with mass m and with a viscous positive damping, a linear stiffness and a harmonic excitation. The parameters have the following physical meaning: c—damping $(2h = c/m);k$—stiffness $(\alpha _{0}^{2} = k/m);P_{0}$—amplitude of exciting force $(q = p_{0}/m)$ and ω is frequency of excitation. The characteristic equation is

$$\displaystyle{r^{2} + 2hr +\alpha _{ 0}^{2} = 0,}$$

which gives the roots $r_{1,2} = -h \pm \sqrt{h^{2 } -\alpha _{ 0 }^{2}}$. If h > α ₀, then we have two real roots and oscillations do not appear. If h = α ₀ we have so-called critical damping $c_{cr} = 2\sqrt{km}$ and the root is double. Here we consider the case $\alpha _{0}^{2} > h^{2}$ which corresponds to oscillations and $r_{1,2} = -h \pm i\lambda$, where $\lambda = \sqrt{\alpha _{0 }^{2 } - h^{2}}$. Since $e^{(-h+i\lambda )t} = e^{-ht}(\cos \lambda t + i\sin \lambda t)$ and according to our earlier discussion we can take the following real-valued general solution corresponding to homogeneous equation: $y = C_{1}e^{-ht}\cos \lambda t + C_{2}e^{-ht}\sin \lambda t$. As it has been mentioned, the form of the particular solution to the non-homogeneous equation is similar to the right-hand side excitation and is assumed to be: $y_{p} = A\cos \omega t + B\sin \omega t$. Since

$$\displaystyle\begin{array}{rcl} & y^{{\prime}} = -A\omega \sin \omega t + B\omega \cos \omega t, & {}\\ & y^{{\prime\prime}} = -A\omega ^{2}\cos \omega t - B\omega ^{2}\sin \omega t,& {}\\ \end{array}$$

hence

$$\displaystyle\begin{array}{rcl} L(y_{p})& =& -A\omega ^{2}\cos \omega t-B\omega ^{2}\sin \omega t-2hA\omega \sin \omega t+2hB\omega \cos \omega t +\alpha _{ 0}^{2}A\cos \omega t {}\\ & & +\alpha _{0}^{2}B\sin \omega t = q\cos \omega t. {}\\ \end{array}$$

Equating the terms standing by sinω t and cosω t yields

$$\displaystyle\begin{array}{rcl} & A(\alpha _{0}^{2} -\omega ^{2}) + 2h\omega B = q, & {}\\ & -2h\omega A + (\alpha _{0}^{2} -\omega ^{2})B = 0,& {}\\ \end{array}$$

and therefore

$$\displaystyle\begin{array}{rcl} & A = \frac{W_{A}} {W} = \frac{\left \vert \begin{array}{*{10}c} q& 2h\omega \\ 0&\alpha _{0}^{2} -\omega ^{2} \end{array} \right \vert } {\left \vert \begin{array}{*{10}c} \alpha _{0}^{2} -\omega ^{2} & 2h\omega & \\ &-2h\omega &\alpha _{0}^{2} -\omega ^{2} \end{array} \right \vert } = \frac{q(\alpha _{0}^{2}-\omega ^{2})} {(\alpha _{0}^{2}-\omega ^{2})^{2}+4h^{2}\omega ^{2}},& {}\\ & B = \frac{W_{B}} {W} = \frac{\left \vert \begin{array}{*{10}c} \alpha _{0}^{2} -\omega ^{2} & q \\ -2h\omega &0\\ \end{array} \right \vert } {\left \vert \begin{array}{*{10}c} \alpha _{0}^{2} -\omega ^{2} & 2h\omega \\ -2h\omega &\alpha _{0}^{2} -\omega ^{2}\\ \end{array} \right \vert } = \frac{2h\omega q} {(\alpha _{0}^{2}-\omega ^{2})^{2}+4h^{2}\omega ^{2}}.& {}\\ \end{array}$$

Finally, the general solution to homogeneous equation has the form

$$\displaystyle{y = C_{1}e^{-ht}\cos \lambda t + C_{ 2}e^{-ht}\sin \lambda t + A\cos \omega t + B\sin \omega t.}$$

 □ 

The obtained solution possesses clear physical interpretation. It is composed of oscillation of autonomous damped system (this part vanishes, when t → +∞) and a particular solution generated by the harmonic excitation. A steady state oscillation is represented by the solution

$$\displaystyle{\mathop{\lim }\limits _{t\rightarrow \infty }y(t) = A\cos \omega t + B\sin \omega t = a\cos (\omega t+\varphi ).}$$

Observe that

$$\displaystyle{a\cos (\omega t+\varphi ) = a\cos \varphi \cos \omega t - a\sin \varphi \sin \omega t = A\cos \omega t + B\sin \omega t.}$$

Equating terms standing by cosω t and sinω t we obtain

$$\displaystyle{A = a\cos \varphi,\quad B = -a\sin \varphi,}$$

and hence

$$\displaystyle\begin{array}{rcl} & a = \sqrt{A^{2 } + B^{2}} = \frac{q} {\sqrt{(\alpha _{0 }^{2 }-\omega ^{2 } ) ^{2 } +4h^{2 } \omega ^{2}}},& {}\\ & \varphi = -\mathrm{arctan}\frac{B} {A} = -\mathrm{arctan} \frac{2h\omega } {\alpha _{0}^{2}-\omega ^{2}}. & {}\\ & & {}\\ \end{array}$$

This result indicates that the oscillations are harmonic with the amplitude a and are delayed by the phase shift ϕ in comparison to excitation. The latter example allows to solve more general problem, which possesses many applications in engineering, and particularly in mechanics.

Consider now the following non-homogeneous problem

$$\displaystyle{ L_{n}(y) = e^{at}[P(t)\cos bt + Q(t)\sin bt], }$$

(5.36)

where a, b ∈ R and P(t) and Q(t) are polynomials of orders p and q, respectively. Let the characteristic equation corresponding to the homogeneous equation has the form

$$\displaystyle{ r^{n} + p_{ 1}r^{n-1} + \cdots + p_{ n} = 0. }$$

(5.37)

By s we denote the largest values among p and q (for p = q we take $s = p = q$). If a + ib is not a solution to the characteristic equation (5.38), then also a − ib is not a solution. In this case the being sought particular solution has the form

$$\displaystyle{y_{p} = e^{at}[R(t)\cos bt + S(t)\sin bt],}$$

where R(t) and S(t) are polynomials of an order not higher than s. If a + ib is repeated root of multiplicity k, then

$$\displaystyle{ y_{p} = t^{k}e^{at}[R(t)\cos bt + S(t)\sin bt], }$$

(5.38)

where R(t) and S(t) are polynomials of an order not higher than s.

The steps of the method of undetermined coefficients can be even more detailed presented by introducing the following theorem (see [191]).

Theorem 5.15.

Apply the following steps to solve L _n (y) = f(t) using the method of undetermined coefficients:

• (i) Factor the characteristic polynomial $w_{L_{n}}(r)$ of L _n into linear and irreducible quadratic factors and form a basis B _L for the kernel of L.

• (ii) Separate f(t) into groups of terms that are annihilated by a single annihilator. Form w _A (r), the characteristic polynomial for an annihilator of f(t) factored into linear and irreducible quadratic factors. Recall that for a given function f, an annihilator of f is a linear operator A such that A(f) = 0. For instance, $y^{{\prime\prime}} + 16y$ is an annihilator of sin ϕt and cos ϕt, because the characteristic equation is $(r^{2} + 16) = (r + 4i)(r - 4i).$

• (iii) The characteristic polynomial of A _L is defined by $w_{A}(r)w_{L}(r)$ . Form the basis B _AL , and find B consisting of the functions in B _AL that are not in B _L .

• (iv) Form y _p (t) as a linear combination of the functions in B and equate coefficients of L(y _p ) = f(t).

• (v) Determine values for the coefficients and define a particular solution.

• (vi) Find $y(t) = y_{g}(t) + y_{p}(t).$

Example 5.7.

Solve the differential equation $L(y) = y^{{\prime\prime\prime}}- 2y^{{\prime\prime}} + 9y^{{\prime}}- 18y = e^{t} + 3\sin 2t$ using the method of undetermined coefficients.

• (i) The characteristic polynomial $w_{L}(r) = r^{3} - 2r^{2} + 9r - 18 = (r - 2)(r^{2} + 9)$, and hence $\{B_{L} = e^{2t},\sin 3t,\cos 3t\}$.

• (ii) The function $f(t) = e^{t} + 3\sin 2t$ is annihilated by the characteristic polynomial $w_{A} = (r - 1)(r^{2} + 4)$, and the basis for kernel of A is $B_{A} =\{ e^{t},\sin 2t,\cos 2t\}$.

• (iii) The characteristic polynomial of A _L is $w_{A}w_{L} = (r - 2)(r^{2} + 9)(r - 1)(r^{2} + 4)$, and the kernel of A _L is $B_{AL} =\{ e^{t},e^{2t},\sin 2t,\cos 2t,\sin 3t,\cos 3t\}$. We have found also that $B = B_{A} =\{ e^{t},\sin 2t,\cos 2t\}.$

• (iv) The particular solution $y_{p} = ae^{t} + b\sin 2t + c\cos 2t$. The successive differentiation gives

  $$\displaystyle\begin{array}{rcl} & y^{{\prime}}_{p} = ae^{t} + 2b\cos 2t - 2c\sin 2t, & {}\\ & y^{{\prime\prime}}_{p} = ae^{t} - 4b\sin 2t - 4c\cos 2t,& {}\\ & y^{{\prime\prime\prime}}_{p} = ae^{t} - 8b\sin 2t + 8c\cos 2t,& {}\\ \end{array}$$

  and

  $$\displaystyle\begin{array}{rcl} L(y_{p})& =& ae^{t} - 8b\sin 2t + 8c\cos 2t - 2(ae^{t} - 4b\sin 2t + 4c\cos 2t) {}\\ & & +9(ae^{t} + 2b\cos 2t - 2c\sin 2t) - 18(ae^{t} + b\sin 2t + c\cos 2t) {}\\ & =& -10ae^{t} - 18(b + c)\sin 2t + 2(9b - c)\cos 2t. {}\\ \end{array}$$

  Equating the appropriate coefficients with those of f(t) gives

  $$\displaystyle{\begin{array}{*{10}c} e^{t}: & -10a = 1, \\ \sin 2t:&-18(b + c) = 3, \\ \cos 2t:& 9b - c = 0.\\ \end{array} }$$

• (v) The solution of the latter linear algebraic equation is: $a = -10,b = -\frac{1} {60}$, $c = -\frac{3} {20}$, and hence the particular solution is

  $$\displaystyle{y_{p}(t) = -10e^{t} - \frac{1} {60}\sin 2t - \frac{3} {20}\cos 2t.}$$

• (vi) Finally, the general solution is

  $$\displaystyle{y = C_{1}e^{2t} + C_{ 2}\sin 3t + C_{3}\cos 3t - 10e^{t} - \frac{1} {60}\sin 2t - \frac{3} {20}\cos 2t.}$$

 □ 

Now we briefly describe the method of variation of parameters. Recall that even in the method of undetermined coefficients we have found the kernel of a homogeneous equation L _n (y) = 0 first.

Theorem 5.16.

A particular solution y _p (t) of non-homogeneous equation L _n (y) = f(t) can be found by quadratures if it will be sought in the form

$$\displaystyle{ y_{p}(t) = C_{1}(t)y_{1}(t) + C_{2}(t)y_{2}(t) + \cdots + C_{n}(t)y_{n}(t), }$$

(5.39)

where $\{y_{1}(t),y_{2}(t),\ldots,y_{n}(t)\}$ is the basis of the kernel of L _n = 0.

Theorem 5.17.

If the functions $C_{1}(t),\ldots,C_{n}(t)$ satisfy the system of equations

$$\displaystyle\begin{array}{rcl} & C^{{\prime}}_{1}y_{1} + C^{{\prime}}_{2}y_{2} + \cdots + C^{{\prime}}_{n}y_{n} & \\ & C^{{\prime}}_{1}y^{{\prime}}_{1} + C^{{\prime}}_{2}y^{{\prime}}_{2} + \cdots + C^{{\prime}}_{n}y^{{\prime}}_{n} & \\ & \vdots & \\ & C^{{\prime}}_{1}y_{1}^{(n-1)} + C^{{\prime}}_{2}y_{2}^{(n-1)} + \cdots + C^{{\prime}}_{n}y_{n}^{(n-1)},&{}\end{array}$$

(5.40)

then the function y _p defined by  (5.39) satisfies the non-homogeneous equation L _n (y) = f(t).

Remark 5.4.

The function y _p (t) possesses the form of a general integral of L _n (y) = 0, where now the constant are the functions depending on t.

Remark 5.5.

Observe that $C_{1}^{{\prime}},C_{2}^{{\prime}},\ldots,C_{n}^{{\prime}}$ are uniquely defined, since the determinant of the system (5.41) is the Wronskian determinant, and by definition it never equals zero.

In order to get $C_{1}(t),\ldots,C_{n}(t)$ we integrate the functions $C_{1}^{{\prime}}(t),\ldots,C_{n}^{{\prime}}(t)$ without the constant of integrations, because they appear already in the part of n general solutions of the homogeneous equation.

Example 5.8.

Find a general solution of the equation

$$\displaystyle{L_{2}(y) = y^{{\prime\prime}} + y = t^{2}.}$$

The kernel of L ₂(y) = 0 is {sint, cost}. The general solution of $L_{2}(y) = t^{2}$ is $y = y_{p} + C_{1}\cos t + C_{2}\sin t$.

The functions y _p (t) is sought in the form

$$\displaystyle{y_{p}(t) = C_{1}(t)\cos t + C_{2}(t)\sin t.}$$

From (5.41) we obtain

$$\displaystyle\begin{array}{rcl} & C^{{\prime}}_{1}\cos t + C^{{\prime}}_{2}\sin t = 0, & {}\\ & -C^{{\prime}}_{1}\cos t + C^{{\prime}}_{2}\sin t = t^{2},& {}\\ \end{array}$$

and hence

$$\displaystyle{C_{1}^{{\prime}} = -t^{2}\sin t,\quad C_{ 2}^{{\prime}} = t^{2}\cos t.}$$

Integration of $C_{1}^{{\prime}},C_{2}^{{\prime}}$ yields

$$\displaystyle{C_{1}(t) = (t^{2} - 2)\cos t - 2t\sin t,\quad C_{ 2}(t) = (t^{2} - 2)\sin t + 2t\cos t}$$

and

$$\displaystyle{y = t^{2} - 2 + C_{ 1}\cos t + C_{2}\sin t.}$$

5.5 Differential Equations with Variable Coefficients

We begin with a special class of higher order homogeneous differential equations, which can be reduced to the differential equations with constant coefficients. A solution to the differential equation

$$\displaystyle{ L_{n}(y) = p_{n}t^{n}\frac{d^{n}y} {dt^{n}} + p_{n-1}t^{n-1}\frac{d^{n-1}y} {dt^{n-1}} + \cdots + p_{1}t\frac{dy} {dt} + p_{0}y, }$$

(5.41)

known as the Cauchy–Euler differential equation, can be sought in the form

$$\displaystyle{ y = t^{r}. }$$

(5.42)

Substituting (5.42) to (5.41) gives

$$\displaystyle{ L_{n}(t^{r}) = w(r)t^{r}, }$$

(5.43)

and hence L _n  = 0, when w(r) = 0. The analysis is similar to previous one, but now instead of the function e ^rt, we have t ^r.

• (i) Assume that there are $r_{1},r_{2},\ldots,r_{m}$ distinct real roots (m ≤ n), then $t^{r_{1}},t^{r_{2}},\ldots,t^{r_{m}}$ are real independent solutions, and

  $$\displaystyle{ y(t) = C_{1}t^{r_{1} } + C_{2}^{r_{2} } + \cdots + C_{m}t^{r_{m} } }$$

  (5.44)

  is a general solution (it defines an m-dimensional subspace of the kernel of (5.41)).

• (ii) Assume that r ₀ is a root with kth multiplicity, then its subkernel follows

  $$\displaystyle{\{t^{r_{0} },t^{r_{0} }\ln t,\ldots,t^{r_{0} }(\ln t)^{k-1}\},}$$

  and a general solution

  $$\displaystyle{ y(t) = C_{1}t^{r_{0} } + C_{2}t^{r_{0} }\ln t + \cdots + C_{k}t^{r_{0} }(\ln t)^{k-1} }$$

  (5.45)

  defines a k-dimensional subspace.

• (iii) Assume that r = a ± bi, then t ^acos(blnt) and t ^asin(blnt) are real linearly independent solutions. In addition, if a ± bi has multiplicity k, then

  $$\displaystyle{ \begin{array}{*{10}c} t^{a}\cos (b\ln t),t^{a}\ln t\cos (b\ln t),\ldots,t^{a}(\ln t)^{k-1}\cos (b\ln t), \\ t^{a}\sin (b\ln t),t^{a}\ln t\sin (b\ln t),\ldots,t^{a}(\ln t)^{k-1}\sin (b\ln t),\\ \end{array} }$$

  (5.46)

  are 2k real linearly independent solutions in the kernel of (5.41). Their linear combinations span a 2k-dimensional subspace.

Now we show how to reduce the homogeneous and non-homogeneous Cauchy–Euler equation to an equation with constant coefficients.

Theorem 5.18.

Introducing a variable x = ln t in the non-homogeneous Cauchy–Euler equation

$$\displaystyle{ L_{n}(y) = p_{n}t^{n}\frac{d^{n}y} {dt^{n}} + p_{n-1}t^{n-1}\frac{d^{n-1}y} {dt^{n-1}} + \cdots + p_{1}t\frac{dy} {dt} + p_{0}y = f(t), }$$

(5.47)

leads to conversion to the differential equation with constant coefficients

$$\displaystyle{ L[y(e^{x})] = g(x), }$$

(5.48)

where g(x) = f(e ^x ).

Example 5.9.

Transform the Cauchy–Euler differential equation

$$\displaystyle{t^{3}y^{{\prime\prime\prime}}- 4t^{2}y^{{\prime\prime}} + 3ty^{{\prime}} + y = 0}$$

to a differential equation with constant coefficients.

Let t + e ^x, and use the chain rule to get

$$\displaystyle\begin{array}{rcl} & \frac{dy} {dt} = \frac{dy} {dx} \frac{dx} {dt} = \frac{dy} {dx} \frac{1} {t},\quad \mathrm{i.e.}\quad t\frac{dy} {dt} = \frac{dy} {dx}; & {}\\ & \frac{d^{2}y} {dt^{2}} = \frac{d} {dt}\left (\frac{dy} {dt} \right ) = \frac{d} {dt}\left (\frac{dy} {dx} \frac{1} {t} \right ) = \frac{d} {dt}\left (\frac{dy} {dx}\right )\frac{1} {t} + \frac{dy} {dx} \frac{d} {dt}\left (\frac{1} {t} \right ) & {}\\ & = \frac{1} {t^{2}} \left (\frac{d^{2}y} {dx^{2}} -\frac{dy} {dx}\right )\quad \mathrm{i.e.}\quad t^{2}\frac{d^{2}y} {dt^{2}} = \frac{d^{2}y} {dx^{2}} -\frac{dy} {dx}; & {}\\ & \frac{d^{3}y} {dt^{3}} = \frac{d} {dt}\left (\frac{d^{2}y} {dt^{2}} \right ) = \frac{d} {dt}\left [ \frac{1} {t^{2}} \left (\frac{d^{2}y} {dx^{2}} -\frac{dy} {dx}\right )\right ] & {}\\ & = \frac{-2} {t^{3}} \left (\frac{d^{2}y} {dx^{2}} -\frac{dy} {dx}\right ) + \frac{1} {t^{2}} \left [ \frac{d} {dt}\left (\frac{d^{2}y} {dx^{2}} \right ) - \frac{d} {dt}\left (\frac{dy} {dx}\right )\right ] & {}\\ & = \frac{-2} {t^{3}} \left (\frac{d^{2}y} {dx^{2}} -\frac{dy} {dx}\right ) + \frac{1} {t^{3}} \left (\frac{d^{3}y} {dx^{3}} -\frac{d^{2}y} {dx} \right ) & {}\\ & = \frac{1} {t^{3}} \left (\frac{d^{3}y} {dx^{3}} - 3\frac{d^{2}y} {dx^{2}} - 2\frac{dy} {dx}\right ),\quad \mathrm{i.e.}\quad \frac{d^{3}y} {dt^{3}} = \frac{d^{3}y} {dx^{3}} - 3\frac{d^{2}y} {dx^{2}} - 2\frac{dy} {dx}.& {}\\ \end{array}$$

Our investigated equation becomes

$$\displaystyle{\frac{d^{3}y} {dx^{3}} - 7\frac{d^{2}y} {dx^{2}} + 5\frac{dy} {dx} + y = 0.}$$

The characteristic polynomial is

$$\displaystyle{r^{3} - 7r^{2} + 5r + 1 = (r - 1)(r^{2} - 6r - 1)}$$

and possesses the characteristic roots

$$\displaystyle{r_{2} = 1,\quad r_{1} = 3 -\sqrt{10},\quad r_{3} = 3 + \sqrt{10}.}$$

Hence the solution is

$$\displaystyle{y = C_{1}e^{r_{1} x} + C_{ 2}e^{r_{2} x} + C_{ 3}e^{r_{3} x} = C_{ 1}e^{r_{1} \ln t} + C_{ 2}e^{r_{2} \ln t} + C_{ 3}e^{r_{3} \ln t}.}$$

 □ 

A key role in mechanics plays second-order differential equation, since a second derivative corresponds to acceleration of a moving rigid body with one degree of freedom. It can happen that one of its solutions is known (for example, from theoretical considerations or from an experiment), and a question appears, how to find a second linearly independent solution?

Consider $L_{2}(y) = p_{2}(t)y^{{\prime\prime}} + p_{1}(t)y^{{\prime}} + p_{0}(t)y = 0$, and assume y(t) = v(t)y ₁(t), where y ₁(t) satisfies $L_{2}(y_{1}) = 0$. The assumed form of solution refers to the so-called Bernoulli method, which is often applied to solve first-order non-homogeneous differential equations. The successive differentiation gives

$$\displaystyle\begin{array}{rcl} y^{{\prime}}& =& v^{{\prime}}y_{ 1} + vy^{{\prime}}_{ 1}, \\ y^{{\prime\prime}}& =& v^{{\prime\prime}}y_{ 1} + 2v^{{\prime}}y^{{\prime}}_{ 1} + vy^{{\prime\prime}}_{ 1},{}\end{array}$$

(5.49)

and hence

$$\displaystyle\begin{array}{rcl} L_{2}(vy_{1})& =& p_{2}(t)(v^{{\prime\prime}}y_{ 1} + 2v^{{\prime}}y^{{\prime}}_{ 1} + vy^{{\prime\prime}}_{ 1}) + p_{1}(t)(v^{{\prime}}y_{ 1} + vy^{{\prime}}_{ 1}) + p_{0}(t)vy_{1} \\ & =& v^{{\prime\prime}}(p_{ 2}y_{1}) + v^{{\prime}}(2p_{ 2}y^{{\prime}}_{ 1} + p_{1}y_{1}) + v(p_{2}y^{{\prime\prime}}_{ 1} + p_{1}y^{{\prime}}_{ 1} + p_{0}y_{1}) \\ & =& p_{2}y_{1}v^{{\prime\prime}} + (2p_{ 2}y^{{\prime}}_{ 1} + p_{1}y_{1})v^{{\prime}} = 0. {}\end{array}$$

(5.50)

Denoting v ^′ = w one can solve first-order linear differential equation with regard to w ^′ to give

$$\displaystyle{ w(t) = y_{1}^{-2}e^{-\int p_{1}(t)/p_{2}(t)}dt. }$$

(5.51)

Hence

$$\displaystyle{ v(t) =\int y_{1}^{-2}e^{-\int p_{1}(t)/p_{2}(t)dt}dt, }$$

(5.52)

and the second solution is

$$\displaystyle{ y_{2}(t) = v(t)y_{1}(t) = y_{1}(t)\int y_{1}^{-2}(t)e^{-\int p_{1}(t)/p_{2}(t)dt}dt. }$$

(5.53)