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Magazine Chapter 4: Problem 5
$$w^{\prime \prime}+4 w^{\prime}+6 w=0$$
Short Answer
Expert verified
The solution of the differential equation is \(w = e^{-2t}(c1 cos(2t) + c2 sin(2t))\).
Step by step solution
01
Create the Characteristic Equation
Make a characteristic equation using the coefficients of \(w''\), \(w'\), and \(w\) in our differential equation. That gives us the characteristic equation as \(m^2 + 4m + 6 = 0\).
02
Solve the Characteristic Equation
Solve this quadratic equation to find the roots. The quadratic formula \(m = (-b \pm \sqrt{b^2 - 4ac}) / 2a\) can be used to do this. For our equation, \(a = 1\), \(b = 4\), and \(c = 6\). Using these in the formula we get \(m = (-4 \pm \sqrt{4^2 - 4*1*6}) / 2*1\), which simplifies to \(m = -2 \pm \sqrt{-4}\).
03
Analyze the Roots
Since the term under the square root is negative, the roots will be complex or imaginary, of the form \(a \pm bi\). The square root of -4 is 2i. So, our roots become -2 \pm 2i.
04
Write the General Solution
The general solution of the differential equation will be \(w = e^{at}(c1 cos(bt) + c2 sin(bt))\), where a is the real part of the root, and b is the imaginary part of the root for \(a \pm bi\), and \(c1\) and \(c2\) are arbitrary constants. Thus, the general solution for our case will be \(w = e^{-2t}(c1 cos(2t) + c2 sin(2t))\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
When working with a second order linear homogeneous differential equation, like \( w'' + 4w' + 6w = 0 \), we first need to find the **characteristic equation**. The characteristic equation is obtained by replacing the derivative terms with powers of \( m \). Specifically:
- Replace \( w'' \) with \( m^2 \)
- Replace \( w' \) with \( m \)
- Keep \( w \) as it is, multiplied by its coefficient
Quadratic Formula
To find the roots of the characteristic equation \( m^2 + 4m + 6 = 0 \), we use the **quadratic formula**:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this formula, \( a \), \( b \), and \( c \) are the coefficients from the equation \( am^2 + bm + c = 0 \). By substituting \( a = 1 \), \( b = 4 \), and \( c = 6 \), we compute:
- The discriminant \( b^2 - 4ac = 16 - 24 = -8 \)
- This results in \( m = \frac{-4 \pm \sqrt{-8}}{2} \)
Complex Roots
When solving the characteristic equation, the presence of a negative discriminant (\( -8 \) in our case) indicates the roots are not real, but **complex roots**. In the equation \( m = -2 \pm \sqrt{-4} \), the square root of -4 is 2i, where \( i \) is the imaginary unit. Thus, the roots are:
- \( m = -2 + 2i \)
- \( m = -2 - 2i \)
General Solution
With the complex roots \( m = -2 \pm 2i \), we can write the **general solution** of the differential equation. The typical form for complex roots \( a \pm bi \) is:\[ w(t) = e^{at}(c_1 \cos(bt) + c_2 \sin(bt)) \]Here, \( a \) is the real part of the root, and \( b \) is the coefficient of \( i \). For our solution:
- \( a = -2 \), leading to an exponential factor \( e^{-2t} \)
- \( b = 2 \), bringing in \( \cos(2t) \) and \( \sin(2t) \)
