Chapter 4: Problem 15
\(y^{\prime \prime}-2 y^{\prime}+4 y=0\)
Short Answer
Expert verified
The solution is \(y(x) = e^{x}(C_1 \cos(\sqrt{3}x) + C_2 \sin(\sqrt{3}x))\).
Step by step solution
01
Identify the Type of Equation
The given equation \(y'' - 2y' + 4y = 0\) is a homogeneous linear differential equation with constant coefficients.
02
Write the Characteristic Equation
For a second-order differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation is \(ar^2 + br + c = 0\). Here, \(a = 1\), \(b = -2\), and \(c = 4\), so the characteristic equation is \(r^2 - 2r + 4 = 0\).
03
Solve the Characteristic Equation
Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots of the characteristic equation. For \(r^2 - 2r + 4 = 0\), we have \(r = \frac{2 \pm \sqrt{(-2)^2 - 4 imes 1 imes 4}}{2}\), which simplifies to \(r = \frac{2 \pm \sqrt{4 - 16}}{2}\).
04
Calculate the Discriminant
Calculate the discriminant \(b^2 - 4ac\). Here, it is \(4 - 16 = -12\). Since it is negative, the roots are complex.
05
Find Complex Roots
The roots are given by \(r = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}\). Thus, the roots are complex numbers: \(r_1 = 1 + i\sqrt{3}\) and \(r_2 = 1 - i\sqrt{3}\).
06
Form the General Solution
For complex roots \(r = \alpha \pm i\beta\), the solution is \(y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))\). Here, \(\alpha = 1\) and \(\beta = \sqrt{3}\), so the general solution is \(y(x) = e^{x}(C_1 \cos(\sqrt{3}x) + C_2 \sin(\sqrt{3}x))\).
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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 second-order homogeneous linear differential equations like \(y'' - 2y' + 4y = 0\), a key step is to form the characteristic equation. This is an algebraic equation derived directly from the differential equation itself. The standard form for a second-order equation is \(ay'' + by' + cy = 0\).The characteristic equation corresponding to this is \(ar^2 + br + c = 0\). By substituting the coefficients of the differential equation into this standard form, you replace the derivatives with a polynomial in \(r\), which models the behavior of the solutions.
- For our example: \(a = 1\), \(b = -2\), and \(c = 4\).
- Thus, the characteristic equation becomes \(r^2 - 2r + 4 = 0\).
Complex Roots
The roots of the characteristic equation can tell us a lot about the solutions of the original differential equation. Complex roots arise when the discriminant \(b^2 - 4ac\) is negative, meaning that the square root of a negative number is involved. This gives rise to imaginary numbers.In the characteristic equation \(r^2 - 2r + 4 = 0\):
- The discriminant, \(b^2 - 4ac\), is calculated as \((-2)^2 - 4 \times 1 \times 4 = -12\) which is negative.
- Therefore, our roots are complex: \(r = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}\).
General Solution
Once you have the complex roots of the characteristic equation, you can derive the general solution to the differential equation. For complex roots of the form \(r = \alpha \pm i\beta\), the general solution is expressed using exponential and trigonometric functions:\[ y(x) = e^{\alpha x}\left(C_1 \cos(\beta x) + C_2 \sin(\beta x)\right) \]For the given roots \(1 \pm i\sqrt{3}\), we have:
- \(\alpha = 1\)
- \(\beta = \sqrt{3}\)