Chapter 7: Problem 45
Solve the boundary-value problem, if possible. $$ y^{\prime \prime}+9 y=0, \quad y(0)=4, \quad y\left(\frac{\pi}{3}\right)=-4 $$
Short Answer
Expert verified
The solution is \( y(x) = 4 \cos(3x) - 4\sqrt{3} \sin(3x) \).
Step by step solution
01
Solve the Differential Equation
The given differential equation is \( y'' + 9y = 0 \). We recognize this as a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is \( r^2 + 9 = 0 \), which simplifies to \( r^2 = -9 \). Solving gives \( r = \pm 3i \). This leads to the general solution: \( y(x) = C_1 \cos(3x) + C_2 \sin(3x) \) where \( C_1 \) and \( C_2 \) are constants.
02
Apply the First Boundary Condition
Use the first boundary condition \( y(0) = 4 \). Substituting in the general solution gives \( 4 = C_1 \cos(0) + C_2 \sin(0) \). Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \), we find \( 4 = C_1 \). Thus, \( C_1 = 4 \).
03
Apply the Second Boundary Condition
Now use the second boundary condition \( y\left(\frac{\pi}{3}\right) = -4 \). Substituting into the general solution gives \( -4 = 4 \cos\left(\frac{\pi}{3}\right) + C_2 \sin\left(\frac{\pi}{3}\right) \). We know \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), so substituting these gives \( -4 = 4 \times \frac{1}{2} + C_2 \times \frac{\sqrt{3}}{2} \). Simplifying gives \( -4 = 2 + C_2 \frac{\sqrt{3}}{2} \), which rearranges to \( C_2 \frac{\sqrt{3}}{2} = -6 \), so \( C_2 = -\frac{12}{\sqrt{3}} = -4\sqrt{3} \).
04
Write the Particular Solution
Substitute the values of \( C_1 \) and \( C_2 \) back into the general solution to get the particular solution: \( y(x) = 4 \cos(3x) - 4\sqrt{3} \sin(3x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Differential Equation
A homogeneous differential equation is a special type of equation that has equal terms on both sides when set to zero. In our problem, the given differential equation is \( y'' + 9y = 0 \). This is referred to as a **second-order linear homogeneous differential equation** because:
- It's second-order, meaning it involves the second derivative \( y'' \).
- It's linear, as the equation includes no products or powers of the unknown function \( y \) or its derivatives other than the first power.
- It's homogeneous because all terms are dependent on \( y \) or its derivatives and the equation equals zero.
Characteristic Equation
The characteristic equation is a tool that helps us solve homogeneous differential equations, especially those with constant coefficients. When faced with the equation \( y'' + 9y = 0 \), we derive its characteristic equation as follows:
Write the equation as "characteristic equation" form: let \( y = e^{rx} \). Then, \( y' = re^{rx} \) and \( y'' = r^2e^{rx} \).
Substitute into the differential equation: \( r^2e^{rx} + 9e^{rx} = 0 \).
Factor the common term out: \( e^{rx}(r^2 + 9) = 0 \).
This leads to \( r^2 + 9 = 0 \), our characteristic equation.
By solving \( r^2 + 9 = 0 \), we find \( r = \pm 3i \).
These complex roots indicate that our solutions will involve trigonometric functions:
Write the equation as "characteristic equation" form: let \( y = e^{rx} \). Then, \( y' = re^{rx} \) and \( y'' = r^2e^{rx} \).
Substitute into the differential equation: \( r^2e^{rx} + 9e^{rx} = 0 \).
Factor the common term out: \( e^{rx}(r^2 + 9) = 0 \).
This leads to \( r^2 + 9 = 0 \), our characteristic equation.
By solving \( r^2 + 9 = 0 \), we find \( r = \pm 3i \).
These complex roots indicate that our solutions will involve trigonometric functions:
- \( C_1 \cos(3x) \)
- \( C_2 \sin(3x) \)
Trigonometric Functions
Trigonometric functions play a crucial role in solving differential equations, especially when dealing with complex roots of the characteristic equation. In the boundary-value problem, our general solution involves \( C_1 \cos(3x) + C_2 \sin(3x) \). Here's why these functions are important:
- **Cosine and sine functions** are periodic, meaning they repeat values over regular intervals. This makes them ideal for modeling wave-like behavior in systems.
- The coefficients \( C_1 \) and \( C_2 \) are determined using initial or boundary conditions, such as \( y(0) = 4 \) and \( y\left(\frac{\pi}{3}\right) = -4 \).
- The angles within the functions (e.g., \( 3x \)) result from the complex roots \( \pm 3i \) derived from the characteristic equation.