Chapter 4: Problem 27
\(y^{\prime \prime}-2 y^{\prime}+17 y=0, \quad y(0)=-2, \quad y^{\prime}(0)=3\)
Short Answer
Expert verified
The particular solution is \(y(t) = e^t (-2 \cos(4t) + \frac{3}{4} \sin(4t))\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is a linear second-order homogeneous differential equation with constant coefficients: \[ y'' - 2y' + 17y = 0 \] This equation has characteristic roots that determine the form of the general solution.
02
Find the Characteristic Equation
To solve the differential equation, form the characteristic equation by substituting \(y = e^{rt}\) into the differential equation. This yields: \[ r^2 - 2r + 17 = 0 \]
03
Solve the Characteristic Equation
The characteristic equation \(r^2 - 2r + 17 = 0\) is a quadratic equation. Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -2\), \(c = 17\). Calculate:\[ r = \frac{2 \pm \sqrt{4 - 68}}{2} = 1 \pm i4 \] The roots are complex: \(r = 1 \pm 4i\).
04
Write the General Solution
For complex roots \(r = \alpha \pm \beta i\), where \(\alpha = 1\) and \(\beta = 4\), the general solution is:\[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]Substitute \(\alpha\) and \(\beta\):\[ y(t) = e^{t} (C_1 \cos(4t) + C_2 \sin(4t)) \]
05
Apply Initial Conditions to Find Constants
Use the initial conditions \(y(0) = -2\) and \(y'(0) = 3\) to determine \(C_1\) and \(C_2\). First, substitute \(t = 0\) in the general solution:\[ y(0) = e^{0}(C_1 \cdot 1 + C_2 \cdot 0) = C_1 = -2 \]Next, differentiate the general solution for \(y'(t)\):\[ y'(t) = e^t (C_1 \cos(4t) + C_2 \sin(4t)) + e^t (-4C_1 \sin(4t) + 4C_2 \cos(4t)) \]Substitute \(t = 0\):\[ y'(0) = e^{0}(C_1 \cdot 0 + C_2 \cdot 4) = 4C_2 = 3 \]Solve for \(C_2\): \(C_2 = \frac{3}{4}\).
06
Write the Particular Solution
Substitute \(C_1 = -2\) and \(C_2 = \frac{3}{4}\) back into the general solution to write the particular solution:\[ y(t) = e^t (-2 \cos(4t) + \frac{3}{4} \sin(4t)) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
When dealing with linear second-order homogeneous differential equations like the one presented in our exercise—\( y'' - 2y' + 17y = 0 \)—a key step is to convert it into a characteristic equation. This transformation simplifies the process of finding the solution. With the assumption that the solution takes the form \( y = e^{rt} \), we substitute this into the differential equation. By differentiating, we get \( y'' = r^2 e^{rt} \) and \( y' = re^{rt} \). Substituting these into the original equation leads to:
- \( r^2 e^{rt} - 2re^{rt} + 17e^{rt} = 0 \)
Complex Roots
Finding roots of the characteristic equation \( r^2 - 2r + 17 = 0 \) involves using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = 17 \). Plugging in these values, we calculate:
- \( r = \frac{2 \pm \sqrt{4 - 68}}{2} = 1 \pm 4i \)
- \( y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \)
Homogeneous Differential Equations
Homogeneous differential equations, such as \( y'' - 2y' + 17y = 0 \), play a crucial role in understanding systems governed by linear rules without external forces. In such equations, each term includes the function or its derivatives, and the right-hand side equals zero, indicating that the system is self-contained.These equations are foundational in modeling many physical systems, where initial conditions solely influence behavior over time. For second-order equations, the characteristic equation derived offers a pathway to defining the system's evolution. Depending on the roots, solutions can vary from exponential growth/decay to oscillatory patterns. Working through the solution requires:
- Finding a characteristic equation
- Solving for its roots
- Constructing a general solution based on these roots