The auxiliary equation for the given differential equation has complex roots. Find a general solution. 8y'' -8y' +52y = 0 y(t) =
The auxiliary equation for the given differential equation has complex roots. Find a general solution. 8y'' -8y' +52y = 0 y(t) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Title: Solving Differential Equations with Complex Roots**
To solve the given differential equation with complex roots, we first recognize the provided equation and proceed with finding the general solution.
**Given Differential Equation:**
\[
8y'' - 8y' + 52y = 0
\]
**Step 1: Auxiliary Equation**
The first step is to formulate the corresponding auxiliary equation. Using the form \(ay'' + by' + cy = 0\), we get the auxiliary equation:
\[
8r^2 - 8r + 52 = 0
\]
**Step 2: Solving the Auxiliary Equation**
Next, solve for roots of the quadratic equation \(8r^2 - 8r + 52 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Here, \(a = 8\), \(b = -8\), and \(c = 52\).
Calculate the discriminant:
\[
\Delta = b^2 - 4ac = (-8)^2 - 4(8)(52) = 64 - 1664 = -1600
\]
Since the discriminant \(\Delta\) is less than zero, the roots will be complex.
Calculate \(r\):
\[
r = \frac{-(-8) \pm \sqrt{-1600}}{2 \cdot 8} = \frac{8 \pm 40i}{16} = \frac{1}{2} \pm \frac{5i}{2}
\]
The complex roots are:
\[
r = \frac{1}{2} \pm \frac{5i}{2}
\]
**Step 3: General Solution**
The general solution to the differential equation with complex roots \(r = \alpha \pm \beta i\), where \(\alpha = \frac{1}{2}\) and \(\beta = \frac{5}{2}\), is:
\[
y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right)
\]
Substituting \(\alpha\) and \(\beta\):
\[
y(t) = e^{\frac{1}{2} t} \left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57d9078-3a26-4ab9-a5c6-f67f173c178d%2Fbb3d5c50-a16d-41ab-96c3-b48240f00482%2Fr8yy7m8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Solving Differential Equations with Complex Roots**
To solve the given differential equation with complex roots, we first recognize the provided equation and proceed with finding the general solution.
**Given Differential Equation:**
\[
8y'' - 8y' + 52y = 0
\]
**Step 1: Auxiliary Equation**
The first step is to formulate the corresponding auxiliary equation. Using the form \(ay'' + by' + cy = 0\), we get the auxiliary equation:
\[
8r^2 - 8r + 52 = 0
\]
**Step 2: Solving the Auxiliary Equation**
Next, solve for roots of the quadratic equation \(8r^2 - 8r + 52 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Here, \(a = 8\), \(b = -8\), and \(c = 52\).
Calculate the discriminant:
\[
\Delta = b^2 - 4ac = (-8)^2 - 4(8)(52) = 64 - 1664 = -1600
\]
Since the discriminant \(\Delta\) is less than zero, the roots will be complex.
Calculate \(r\):
\[
r = \frac{-(-8) \pm \sqrt{-1600}}{2 \cdot 8} = \frac{8 \pm 40i}{16} = \frac{1}{2} \pm \frac{5i}{2}
\]
The complex roots are:
\[
r = \frac{1}{2} \pm \frac{5i}{2}
\]
**Step 3: General Solution**
The general solution to the differential equation with complex roots \(r = \alpha \pm \beta i\), where \(\alpha = \frac{1}{2}\) and \(\beta = \frac{5}{2}\), is:
\[
y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right)
\]
Substituting \(\alpha\) and \(\beta\):
\[
y(t) = e^{\frac{1}{2} t} \left(
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