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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)
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.
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
Exercises Exercises
1.
Show why \(x\) is needed in the general solution for repeated roots of the CE
Classify the following differential equations as either homogeneous or nonhomogeneous.
2.
\(\ds 4y'' - 36y' = 0 \)
Find the general solution for each of the following.
3.
\(\ds 4y'' -36y' = 0 \) 4.
\(\ds 4y'' -36y = 0. \) 5.
\(\ds y''- y' - 11y = 0 \) 6.
\(\ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0 \)
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 . Answer
\(\ds y = e^{3t} + 3e^{-3t} \) o \(\ds y = e^{3x} + 3e^{-3x} \)
8.
\(\ds \ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0,\hspace{1cm} \theta(0) = 12, \hspace{1cm} \theta'(0) = -2 \)
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 . Answer \(\ds y = 2e^{t} - e^{-t} \)
Solve the following DEs.
10.
\(\ds w'' + 6w' + 9w = 0 \) 11.
\(\ds m'' = 2m' - m \)
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 . Answer
\(\ds z = (2e^{-2} - e^{-2}x)e^{2x} \) or \(\ds z = (2 - x)e^{2x - 2} \)
Solve the following DEs.
13.
\(\ds y'' + 4y' + 53y = 0 \) 14.
\(\ds z''=-36z \)
Solve the following DEs.
15.
\(\ds y'' = -24y' - 144y \) 16.
\(\ds \frac{d^2w}{dx^2} - 49w = 0 \) 17.
\(\ds \frac{d^2w}{dx^2} + 49w = 0 \) 18.
\(\ds z''- z' - 42z = 0 \)
Find the general solution \(\ds y(t) \) for a linear, homogeneous DEwith constant coefficients which has the given characteristic equation.
19.
\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \) 20.
\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0 \)
Compute the derivative of each of the following functions.
21.
\(\, \ds f(x) = \ln x\cos x \) Answer . 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.
23. Classifying Practice.
24. Classifying Practice.
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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