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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section 6.6 Summary & Exercises

The LHCC chapter (Linear Homogeneous Differential Equations with Constant Coefficients) in "Linear Constant Coefficient Methods" introduces a systematic method to solve higher-order linear differential equations. Here’s a summary based on the content:

Summary of the Key Ideas.

  • Linear Homogeneous Differential Equations with Constant Coefficients (LHCC)
    • These are differential equations where each term consists of a derivative of the unknown function multiplied by a constant.
    • The general form of an LHCC equation is:
      \begin{equation*} a_n\ y^{(n)} + a_{n-1}\ y^{(n-1)} + \dots + a_1\ y' + a_0\ y = 0\text{.} \end{equation*}
  • The Characteristic Equation
    • By assuming a solution of the form \(y = e^{rx}\text{,}\) an LHCC can be reduced to a characteristic polynomial in \(r\text{.}\)
    • The solutions to the characteristic equation determine the form of the general solution.
  • Solution Types
    • Let \(r\) be a solution to the characteristic equation (CE).
    • If \(r\) is different from all other solutions of the CE, then
      \begin{equation*} c e^{r x} \end{equation*}
      is a term of the general solution.
    • If \(r\) is equal to, say, three other solutions of the CE, then
      \begin{equation*} c_1 e^{r x} + c_2 x e^{r x} + c_3 x^2 e^{r x} \end{equation*}
      are terms of the general solution.
    • If \(r = \alpha + i\beta\) or \(r = \alpha - i\beta\text{,}\) then the general solution contains
      \begin{equation*} e^{\alpha x}(c_1\sin(\beta x)+c_2\cos(\beta x))\text{.} \end{equation*}

Method 4. LHCC Method.

The general solution to a linear homogeneous differential equation with constant coefficients (LHCC) of the form
\begin{equation} a_n\ y^{(n)} + a_{n-1}\ y^{(n-1)} + \cdots + a_2\ y'' + a_1\ y' + a_0\ y = 0,\tag{31} \end{equation}
can be found through the following steps...
Step 1: Solve the Characteristic Equation
Solve the characteristic equation (CE)
\begin{equation*} a_n\ r^{n} + a_{n-1}\ r^{n-1} + \cdots + a_2\ r^2 + a_1\ r + a_0 = 0, \end{equation*}
Step 2: Write Down the General Solution
  • Real & Different: \(r_1, r_2, \dots, r_n \)
    \begin{equation*} y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x} + \dots + c_n e^{r_n x}\text{.} \end{equation*}
  • Real & Repeated: \(r_1 \) (multiplicity \(m \))
    \begin{equation*} y(x) = (c_1 + c_2 x + \dots + c_m x^{m-1}) e^{r_1 x}\text{.} \end{equation*}
  • Complex: \(\alpha \pm i\beta \)
    \begin{equation*} y(x) = e^{\alpha x} \left(c_1 \cos(\beta x) + c_2 \sin(\beta x)\right)\text{.} \end{equation*}
  • For mixed root types, combine the corresponding terms to form the complete general solution.

Exercises Exercises

1.

Show why \(x\) is needed in the general solution for repeated roots of the CE
Answer.

Classify the following differential equations as either homogeneous or nonhomogeneous.

2.

\(\ds 4y'' - 36y' = 0 \)
Answer.

Find the general solution for each of the following.

3.

\(\ds 4y'' -36y' = 0 \)
Answer.

4.

\(\ds 4y'' -36y = 0. \)
Answer.

5.

\(\ds y''- y' - 11y = 0 \)
Answer.

6.

\(\ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0 \)
Answer.

Solve the following initial value problems.

7.

\(\ds 4y'' -36y = 0 \hspace{1cm} c_1 \) an \(\ds c_2 \) that satisfy thegiven initial conditions.
Answer.
\(\ds y = e^{3t} + 3e^{-3t} \)\(\ds y = e^{3x} + 3e^{-3x} \)

8.

\(\ds \ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0,\hspace{1cm} \theta(0) = 12, \hspace{1cm} \theta'(0) = -2 \)
Answer.

Boundary Value Problems.

9.

Solve the following boundary value problem.\(\ds y'' - y = 0, \hspace{1cm} y(0) = 1, \hspace{1cm} y(1) = 2e- \frac{1}{e}\)
Answer.
\(\ds y = 2e^{t} - e^{-t} \)

Solve the following DEs.

10.

\(\ds w'' + 6w' + 9w = 0 \)
Answer.

11.

\(\ds m'' = 2m' - m \)
Answer.

Initial Value Problems.

12.

Solve the following initial value problem.\(\ds \frac{d^2z}{dx^2} - 4\frac{dz}{dx} + 4z = 0, \hspace{0.5cm} z(1) = 1, \hspace{0.5cm} z'(1) = 1\)
Answer.
\(\ds z = (2e^{-2} - e^{-2}x)e^{2x} \)\(\ds z = (2 - x)e^{2x - 2} \)

Solve the following DEs.

13.

\(\ds y'' + 4y' + 53y = 0 \)
Answer.

14.

\(\ds z''=-36z \)
Answer.

Solve the following DEs.

15.

\(\ds y'' = -24y' - 144y \)
Answer.

16.

\(\ds \frac{d^2w}{dx^2} - 49w = 0 \)
Answer.

17.

\(\ds \frac{d^2w}{dx^2} + 49w = 0 \)
Answer.

18.

\(\ds z''- z' - 42z = 0 \)
Answer.

Find the general solution \(\ds y(t) \) for a linear, homogeneous DEwith constant coefficients which has the given characteristic equation.

19.

\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)
Answer.

20.

\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0 \)
Answer.

Compute the derivative of each of the following functions.

21.

\(\, \ds f(x) = \ln x\cos x \)
Answer.
\(\ds f'(x) = \left(\frac{1}{x}\right)\cos x + \ln x \left(-\sin x\right) = \frac{\cos x}{x} - \ln x \sin x\)

22. Classifying Practice.

    Select each classification label that applies to the equation
    \begin{equation*} y'' = y' + 6y \end{equation*}
  • Linear
  • Correct, each of the terms are linear.
  • Homogeneous
  • Correct, the constant term is zero.
  • Constant Coefficients
  • Correct, each coefficient is constant.
  • LHCC
  • Correct!

23. Classifying Practice.

    Select each classification label that applies to the equation
    \begin{equation*} 3y''' + y'- \sin(y) = 0 \end{equation*}
  • Linear
  • Incorrect, \(\sin(y)\) is a nonlinear term.
  • Homogeneous
  • Technically, only linear equations can be labeled as homogeneous or not. Since the equation is nonlinear, we do not select it.
  • Constant Coefficients
  • Technically, only linear equations can be labeled as having constant coefficients or not. Since the equation is nonlinear, we do not select it.
  • LHCC
  • Incorrect.

24. Classifying Practice.

    Select each classification label that applies to the equation
    \begin{equation*} y''- 6 = 0 \end{equation*}
  • Linear
  • Correct, both terms are linear.
  • Homogeneous
  • Incorrect, the constant term, \(6\text{,}\) is non-zero.
  • Constant Coefficients
  • Correct, each coefficient is constant.
  • LHCC
  • Incorrect.

25. Classifying Practice.

    Select each classification label that applies to the equation
    \begin{equation*} \frac{d^3y}{dt^3} + k\frac{dy}{dt} = ty, \qquad k \text{ is constant} \end{equation*}
  • Linear
  • Correct, all terms are linear.
  • Homogeneous
  • Correct, the constant term is zero.
  • Constant Coefficients
  • Incorrect, the \(y\) term coefficient, \(t\text{,}\) is not constant.
  • LHCC
  • Incorrect.
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