Chapter 4: Problem 6
In Problems 1-36 find the general solution of the given differential equation. \(3 y^{\prime \prime}+y=0\)
Short Answer
Expert verified
The general solution is \(y(t) = C_1 \cos(\sqrt{\frac{1}{3}}t) + C_2 \sin(\sqrt{\frac{1}{3}}t)\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \(3y'' + y = 0\). Our task is to find the general solution.
02
Write the Characteristic Equation
For the differential equation \(ay'' + by' + cy = 0\), the characteristic equation is formed as \(ar^2 + br + c = 0\). Here, \(a = 3\), \(b = 0\), \(c = 1\). Thus, the characteristic equation becomes:\[3r^2 + 1 = 0\]
03
Solve the Characteristic Equation
Set the characteristic equation to zero and solve for \(r\) to find the roots:\[3r^2 + 1 = 0\]\[3r^2 = -1\]\[r^2 = -\frac{1}{3}\]\[r = \pm i\sqrt{\frac{1}{3}}\]These are complex roots.
04
Form the General Solution with Complex Roots
When the roots of the characteristic equation are complex, \(r = \alpha \pm i\beta\), the general solution is given by:\[y(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t)\]Substitute the real part (\(\alpha = 0\)) and the imaginary part (\(\beta = \sqrt{\frac{1}{3}}\)) into this formula:\[y(t) = C_1 \cos(\sqrt{\frac{1}{3}}t) + C_2 \sin(\sqrt{\frac{1}{3}}t)\]
05
Write the General Solution
The general solution of the given differential equation is:\[y(t) = C_1 \cos(\sqrt{\frac{1}{3}}t) + C_2 \sin(\sqrt{\frac{1}{3}}t)\]where \(C_1\) and \(C_2\) are arbitrary constants.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
The characteristic equation is vital in solving second-order linear homogeneous differential equations. Imagine it as a bridge connecting the differential equation to its solution. For a differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation is \(ar^2 + br + c = 0\). This quadratic equation helps determine the nature of the solutions.
In our exercise, the differential equation is \(3y'' + y = 0\), where \(a = 3\), \(b = 0\), and \(c = 1\). Therefore, the characteristic equation simplifies to:
Remember, the characteristic equation simplifies our task by reducing a differential equation problem to a polynomial problem, which is often more straightforward to solve.
In our exercise, the differential equation is \(3y'' + y = 0\), where \(a = 3\), \(b = 0\), and \(c = 1\). Therefore, the characteristic equation simplifies to:
- \(3r^2 + 1 = 0\)
Remember, the characteristic equation simplifies our task by reducing a differential equation problem to a polynomial problem, which is often more straightforward to solve.
Complex Roots
When solving the characteristic equation \(3r^2 + 1 = 0\), we notice that it involves complex numbers. This happens because the equation results in a negative number under the square root when solving for \(r\):
- Set the equation \(3r^2 = -1\).
- Divide both sides by 3: \(r^2 = -\frac{1}{3}\).
- The roots \(r\) are found by taking the square root, resulting in \(r = \pm i\sqrt{\frac{1}{3}}\).
- \(\alpha = 0\) is the real part.
- \(\beta = \sqrt{\frac{1}{3}}\) is the imaginary part.
General Solution
Upon identifying complex roots, we proceed to form the general solution using a specific formula. For differential equations with roots \( \alpha \pm i\beta \), the general solution is:
\[ y(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t) \]
In our case, \(\alpha = 0\) and \(\beta = \sqrt{\frac{1}{3}}\), simplifying our formula to:
The beauty of this general solution lies in its flexibility. It can model various systems that behave in an oscillatory manner due to their complex roots.
\[ y(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t) \]
In our case, \(\alpha = 0\) and \(\beta = \sqrt{\frac{1}{3}}\), simplifying our formula to:
- Remove \(e^{0t}\) since it equals 1, giving:
- \(y(t) = C_1 \cos(\sqrt{\frac{1}{3}}t) + C_2 \sin(\sqrt{\frac{1}{3}}t)\)
The beauty of this general solution lies in its flexibility. It can model various systems that behave in an oscillatory manner due to their complex roots.