Fundamentals of Second-Order Linear Differential Equations with Constant Coefficients

1Introduction

This lecture derives the solution method for second-order linear differential equations with constant coefficients from linearity and the properties of exponential functions, rather than presenting it merely as a formula.

The standard form is

The assumption that $$a,b,c$$a,b,c are constants is essential. Equations with coefficients depending on $$x$$x,

require a more general framework.

2Homogeneous and Nonhomogeneous Equations

An equation whose right-hand side is zero,

is called a {homogeneous equation}. An equation for which $$f$$f is not identically zero,

is called a {nonhomogeneous equation}. To solve a nonhomogeneous equation, first solve the associated homogeneous equation and then find one particular solution corresponding to the forcing term $$f\left(x\right)$$f(x).

3Proposition 1: Homogeneous Solutions Satisfy Superposition

Suppose that $$y_1,y_2$$y_1,y_2 solve

Then, for arbitrary constants $$C_1,C_2$$C_1,C_2,

is also a solution.

3.1Proof

Define $$L\left[y\right]=a{y}^{\prime }\text{'}+b{y}^{\prime }+cy$$L[y]=ay''+by'+cy. The operator $$L$$L is linear, since

By assumption, $$L\left[y_1\right]=0$$L[y_1]=0 and $$L\left[y_2\right]=0$$L[y_2]=0. Hence

so $$C_1{y}_1+C_2{y}_2$$C_1y_1+C_2y_2 is a homogeneous solution. This property is the {superposition principle}.

4Proposition 2: A Nonhomogeneous Solution Is a Particular Solution Plus a Homogeneous Solution

Suppose that $$y_p$$y_p is one particular solution of

and $$y_h$$y_h solves the associated homogeneous equation

Then $$y=y_p+y_h$$y=y_p+y_h solves the nonhomogeneous equation. Conversely, every solution of the nonhomogeneous equation has this form.

4.1Proof

With $$L\left[y\right]=a{y}^{\prime }\text{'}+b{y}^{\prime }+cy$$L[y]=ay''+by'+cy, linearity gives

Conversely, if $$y$$y is any other solution of the nonhomogeneous equation, then

Thus $$y-y_p$$y-y_p is a homogeneous solution. Writing it as $$y_h$$y_h yields

Equivalently, the solution set of a linear operator equation consists of one particular solution plus every solution of the corresponding homogeneous equation.

5Derivation of the Characteristic Equation

Consider the homogeneous equation

For constant coefficients, differentiation preserves the exponential factor:

It is therefore natural to try $$y=e^{rx}$$y=e^{rx}. Substitution gives

or

Since $$e^{rx}\ne 0$$e^{rx}\ne0, one obtains

This algebraic equation is the {characteristic equation}.

6Justification of the General Solution

The root classification below produces general solutions because the solution space of a second-order linear homogeneous equation is two-dimensional. Once two linearly independent solutions are known, every solution is their linear combination.

In what follows, the scalar field is $$\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$$\mathbb K\in\{\mathbb R,\mathbb C\}. Reduce the equation to the first-order linear system

The map $$u\mapsto Au$$u\mapsto Au is continuous and globally Lipschitz in $$u$$u. The existence and uniqueness theorem therefore gives a unique solution for every initial vector $$u\left(x_0\right)=(\alpha ,\beta {)}^T$$u(x_0)=(\alpha,\beta)^T.

Define the Wronskian of two solutions $$y_1,y_2$$y_1,y_2 by

If $$W(y_1,y_2)\left(x_0\right)\ne 0$$W(y_1,y_2)(x_0)\ne0, the initial-data vectors $$\left(y_1\right(x_0),y_{1^{\prime }}(x_0){)}^T$$(y_1(x_0),y_1'(x_0))^T and $$\left(y_2\right(x_0),y_{2^{\prime }}(x_0){)}^T$$(y_2(x_0),y_2'(x_0))^T are linearly independent. If a linear combination of the two solutions is the zero function, its initial-data vector is also zero; independence of the two vectors forces both coefficients to vanish. Hence $$y_1,y_2$$y_1,y_2 are linearly independent in the solution space.

6.1Proposition 3: Two Independent Solutions Form a Fundamental Set

Let $$a\ne 0$$a\ne0 and consider

Its solution space is two-dimensional. Consequently, if two independent solutions $$y_1,y_2$$y_1,y_2 are found, every solution can be written as

6.2Proof

Since $$a\ne 0$$a\ne0, the equation is equivalent to

Fix a point $$x_0$$x_0 and associate to each solution $$y$$y the initial-data vector

The map $$\mathrm{\Phi }$$\Phi is linear. By the preceding reduction and the existence and uniqueness theorem, for every $$(\alpha ,\beta )$$(\alpha,\beta) there is exactly one solution satisfying

Thus $$\mathrm{\Phi }$$\Phi is a bijection from the solution space to $$K^2$$\mathbb K^2, and the solution space has dimension two. Any two independent solutions therefore form a basis, proving the asserted representation.

Equivalently, the two listed fundamental solutions yield a general solution once their independence is established. Their Wronskians are nonzero: for distinct real roots it is $$(r_2-r_1)e^{(r_1+r_2)x}$$(r_2-r_1)e^{(r_1+r_2)x}; for a repeated root it is $$e^{2rx}$$e^{2rx}; and for the real pair arising from $$\alpha \pm i\beta$$\alpha\pm i\beta it is $$\beta e^{2\alpha x}$$\beta e^{2\alpha x} when $$\beta \ne 0$$\beta\ne0.

7Classification by Characteristic Roots

The fundamental solutions depend on the roots of

8Proposition 4: The Case of Distinct Real Roots

Suppose that the characteristic equation has distinct real roots $$r_1,r_2$$r_1,r_2. Then

is the general homogeneous solution.

8.1Proof

Since each $$r_i$$r_i is a characteristic root,

For $$y_i=e^{r_{i}x}$$y_i=e^{r_ix},

Thus both exponentials are solutions, and Proposition 1 shows that every displayed linear combination is a solution. Their Wronskian is nonzero because $$r_1\ne r_2$$r_1\ne r_2, so Proposition 3 shows that the combination is the general solution. The two constants accommodate the two initial data $$y\left(x_0\right)$$y(x_0) and $$y^{\prime }\left(x_0\right)$$y'(x_0).

9Derivation of the Second Solution $$x{e}^{rx}$$xe^{rx} for a Repeated Root

Suppose the characteristic equation has the repeated root $$\rho$$\rho, so it can be written as

The function $$e^{\rho x}$$e^{\rho x} is one solution, but a second-order equation requires a second independent solution. We therefore test $$x{e}^{\rho x}$$xe^{\rho x}.

9.1Verification

Write $$L=a{D}^2+bD+c$$L=aD^2+bD+c and $$p\left(r\right)=a{r}^2+br+c$$p(r)=ar^2+br+c. A repeated root satisfies

Direct differentiation gives

Therefore

The coefficient of $$e^{\rho x}$$e^{\rho x} is $$2a\rho +b=p^{\prime }\left(\rho \right)$$2a\rho+b=p'(\rho), and that of $$x{e}^{\rho x}$$xe^{\rho x} is $$a{\rho }^2+b\rho +c=p\left(\rho \right)$$a\rho^2+b\rho+c=p(\rho). Both vanish, so

Hence the general solution is

10Complex Roots and Oscillation

Suppose the characteristic roots are

Over the complex numbers, $$e^{(\alpha +i\beta )x}$$e^{(\alpha+i\beta)x} and $$e^{(\alpha -i\beta )x}$$e^{(\alpha-i\beta)x} are solutions. Euler's formula,

provides the real-valued fundamental solutions

If $$\alpha <0$$\alpha<0, their amplitude decays; if $$\alpha =0$$\alpha=0, they describe undamped oscillation. Complex characteristic roots therefore connect directly to oscillations in mechanics and electrical circuits.

11Example 1: Distinct Real Roots

For

the characteristic equation is $$r^2-3r+2=0$$r^2-3r+2=0. Since $$(r-1)(r-2)=0$$(r-1)(r-2)=0, the roots are 1 and 2, and

12Example 2: A Repeated Root

For

the characteristic equation is $$(r-1{)}^2=0$$(r-1)^2=0. Hence

13Example 3: Complex Roots

For

the equation $$r^2+4=0$$r^2+4=0 gives $$r=\pm 2i$$r=\pm2i. The real-valued general solution is therefore

14Extension to Nonhomogeneous Equations

For

Proposition 2 gives

where $$y_h$$y_h is the general solution of the associated homogeneous equation and $$y_p$$y_p is one particular solution. When $$f\left(x\right)$$f(x) is an exponential, trigonometric function, or polynomial, the method of undetermined coefficients may provide $$y_p$$y_p. Variation of parameters applies more generally.

15Connection to the Operator Method

With

the equation becomes

Defining $$P\left(D\right)=a{D}^2+bD+c$$P(D)=aD^2+bD+c gives $$P\left(D\right)y=f\left(x\right)$$P(D)y=f(x). From this perspective, the homogeneous equation $$P\left(D\right)y=0$$P(D)y=0 asks for the kernel of a linear operator. The nonhomogeneous problem asks whether $$f$$f belongs to its image and, when a solution exists, forms the general solution by adding the kernel to one particular solution.

At an introductory level, one should first master the characteristic-equation method. The operator method then provides a theoretical framework that abstracts its algebraic structure.

16Summary

Apply the following solution procedure to a second-order linear equation with constant coefficients.

1. Determine whether the equation is homogeneous or nonhomogeneous.
2. For the homogeneous equation, try $$e^{rx}$$e^{rx} and form the characteristic equation.
3. Classify its roots as distinct real roots, a repeated root, or a complex-conjugate pair.
4. For a nonhomogeneous equation, decompose the solution as $$y=y_h+y_p$$y=y_h+y_p.
5. Finally, use initial or boundary conditions to determine the arbitrary constants.

17Previous Lecture

18Next Lecture

Having established the characteristic-root method for constant coefficients, proceed to the general theory of second-order linear equations with variable coefficients.

19Exercises