3. Solve y" – 2y' + 5y = 0, y(0) = 5, y'(0) = -3.
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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Using Laplace transforms
![**Problem 3:** Solve the differential equation \( y'' - 2y' + 5y = 0 \) with the initial conditions \( y(0) = 5 \) and \( y'(0) = -3 \).
This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we will follow these steps:
1. **Characteristic Equation:** Start by finding the characteristic equation associated with the differential equation. The characteristic equation is obtained by replacing \( y \) with \( e^{rt} \), leading to:
\[ r^2 - 2r + 5 = 0 \]
2. **Solving the Characteristic Equation:** Solve the quadratic equation for \( r \):
- The roots can be found using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For this equation, \( a = 1 \), \( b = -2 \), and \( c = 5 \). Plug these values into the formula:
\[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} \]
\[ r = \frac{2 \pm 4i}{2} = 1 \pm 2i \]
3. **General Solution:** Since the roots are complex, the solution to the differential equation is of the form:
\[ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]
where \( \alpha = 1 \) and \( \beta = 2 \). Therefore, the general solution is:
\[ y(t) = e^{t}(C_1 \cos(2t) + C_2 \sin(2t)) \]
4. **Applying Initial Conditions:** Use the initial conditions to find \( C_1 \) and \( C_2 \):
- \( y(0) = 5 \):
\[ e^{0}(C_1 \cos(0) + C_2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9cba6163-1bf7-4503-a8f6-4b3c743b9de1%2F91b9b9bb-d046-4a8e-8206-b1af0db334dc%2Fvoxzfk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 3:** Solve the differential equation \( y'' - 2y' + 5y = 0 \) with the initial conditions \( y(0) = 5 \) and \( y'(0) = -3 \).
This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we will follow these steps:
1. **Characteristic Equation:** Start by finding the characteristic equation associated with the differential equation. The characteristic equation is obtained by replacing \( y \) with \( e^{rt} \), leading to:
\[ r^2 - 2r + 5 = 0 \]
2. **Solving the Characteristic Equation:** Solve the quadratic equation for \( r \):
- The roots can be found using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For this equation, \( a = 1 \), \( b = -2 \), and \( c = 5 \). Plug these values into the formula:
\[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} \]
\[ r = \frac{2 \pm 4i}{2} = 1 \pm 2i \]
3. **General Solution:** Since the roots are complex, the solution to the differential equation is of the form:
\[ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]
where \( \alpha = 1 \) and \( \beta = 2 \). Therefore, the general solution is:
\[ y(t) = e^{t}(C_1 \cos(2t) + C_2 \sin(2t)) \]
4. **Applying Initial Conditions:** Use the initial conditions to find \( C_1 \) and \( C_2 \):
- \( y(0) = 5 \):
\[ e^{0}(C_1 \cos(0) + C_2
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