Chapter 7: Problem 3
Find the general solution except when the exercise stipulates otherwise. $$\left(D^{2}-2 D+2\right) y=0$$
Short Answer
Expert verified
The general solution is \( y(x) = e^{x} (C_1 \cos{x} + C_2 \sin{x}) \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is a linear homogeneous second-order differential equation with constant coefficients: \( (D^2 - 2D + 2) y = 0 \), where \( D \) represents the differential operator \( \frac{d}{dx} \).
02
Form the Auxiliary Equation
Convert the differential operator equation into its corresponding characteristic equation by replacing \( D \) with \( r \): \( r^2 - 2r + 2 = 0 \).
03
Solve the Auxiliary Equation
The characteristic equation \( r^2 - 2r + 2 = 0 \) is quadratic. Use the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2, \) and \( c = 2 \). Substitute these values to obtain:\[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} \]This simplifies to \( r = 1 \pm i \), indicating complex roots.
04
Write the General Solution
For complex roots \( r = \alpha \pm \beta i \), the general solution is given by:\[ y(x) = e^{\alpha x} (C_1 \cos{\beta x} + C_2 \sin{\beta x}) \]In this case, \( \alpha = 1 \) and \( \beta = 1 \), so the general solution is:\[ y(x) = e^{x} (C_1 \cos{x} + C_2 \sin{x}) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
In order to solve a second-order differential equation with constant coefficients, we rely on converting it into an algebraic equation called the **characteristic equation**. This transformation makes it easier to find solutions, as we switch variables from derivatives to algebraic terms. To form the characteristic equation, replace the differential operator \(D\) with \(r\). So, a differential equation like \((D^2 - 2D + 2)y = 0\) becomes \(r^2 - 2r + 2 = 0\). Breaking down the process this way:
- Replace \(D\) with \(r\) as if it's an algebraic expression
- This step effectively helps you find the roots, which leads to the general solution
- Think of the characteristic equation as the bridge from calculus to algebra
Complex Roots
Sometimes, the roots of the characteristic equation are not real numbers; they are complex. Complex roots occur when the discriminant in the quadratic formula (\(b^2 - 4ac\)) is negative. In the characteristic equation \(r^2 - 2r + 2 = 0\), you find: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{-4}}{2} \]This results in the complex roots \(r = 1 \pm i\). When you encounter such roots, it implies oscillatory behavior and exponential growth in solutions:
- The real part of the root, \(\alpha\), informs the exponential growth/decay
- The imaginary part, \(\beta\), relates to the oscillations (think sine and cosine functions)
- Complex roots typically indicate a phase shift or rotation in the plane of solutions
General Solution
Once we have the roots (whether real or complex), we can write the general solution to the second-order differential equation. For complex roots \( r = \alpha \pm \beta i \), the general solution takes the form:\[ y(x) = e^{\alpha x} (C_1 \cos{\beta x} + C_2 \sin{\beta x}) \]The general solution is a blueprint for describing all possible solutions:
- \(e^{\alpha x}\) represents the exponential growth or decay part
- \(C_1\) and \(C_2\) are constants determined by initial conditions
- \(\cos\) and \(\sin\) functions describe the oscillatory nature