Solve the given initial value problem. y"-10y' +25y = 0; The solution is y(t) = y(0)=-3, y'(0) = - 43 3
Solve the given initial value problem. y"-10y' +25y = 0; The solution is y(t) = y(0)=-3, y'(0) = - 43 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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![### Solving the Initial Value Problem
**Problem Statement:**
Solve the given initial value problem:
\[ y'' - 10y' + 25y = 0 \]
\[ y(0) = -3 \]
\[ y'(0) = -\frac{43}{3} \]
**Solution:**
To solve the second-order linear differential equation with constant coefficients, we follow these steps:
1. **Find the characteristic equation:**
The characteristic equation for the differential equation \( y'' - 10y' + 25y = 0 \) is formed by assuming solutions of the form \( y = e^{rt} \). Substituting this into the differential equation leads to:
\[ r^2 - 10r + 25 = 0 \]
2. **Solve the characteristic equation:**
This is a quadratic equation. We can solve it using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -10 \), and \( c = 25 \). Substituting these values in, we get:
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(25)}}{2(1)} \]
\[ r = \frac{10 \pm \sqrt{100 - 100}}{2} \]
\[ r = \frac{10 \pm \sqrt{0}}{2} \]
\[ r = \frac{10}{2} \]
\[ r = 5 \]
This root has a multiplicity of 2 (repeated root).
3. **Form the general solution:**
When we have a repeated root \( r = 5 \), the general solution to the differential equation is of the form:
\[ y(t) = (C_1 + C_2 t) e^{5t} \]
4. **Apply initial conditions:**
To find \( C_1 \) and \( C_2 \), we use the initial conditions \( y(0) = -3 \) and \( y'(0) = -\frac{43}{3} \).
- Substituting \( t = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57d9078-3a26-4ab9-a5c6-f67f173c178d%2Fbdf7d19a-97bc-4fd5-ada9-1ec735ae950c%2Fhvvjkm8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving the Initial Value Problem
**Problem Statement:**
Solve the given initial value problem:
\[ y'' - 10y' + 25y = 0 \]
\[ y(0) = -3 \]
\[ y'(0) = -\frac{43}{3} \]
**Solution:**
To solve the second-order linear differential equation with constant coefficients, we follow these steps:
1. **Find the characteristic equation:**
The characteristic equation for the differential equation \( y'' - 10y' + 25y = 0 \) is formed by assuming solutions of the form \( y = e^{rt} \). Substituting this into the differential equation leads to:
\[ r^2 - 10r + 25 = 0 \]
2. **Solve the characteristic equation:**
This is a quadratic equation. We can solve it using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -10 \), and \( c = 25 \). Substituting these values in, we get:
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(25)}}{2(1)} \]
\[ r = \frac{10 \pm \sqrt{100 - 100}}{2} \]
\[ r = \frac{10 \pm \sqrt{0}}{2} \]
\[ r = \frac{10}{2} \]
\[ r = 5 \]
This root has a multiplicity of 2 (repeated root).
3. **Form the general solution:**
When we have a repeated root \( r = 5 \), the general solution to the differential equation is of the form:
\[ y(t) = (C_1 + C_2 t) e^{5t} \]
4. **Apply initial conditions:**
To find \( C_1 \) and \( C_2 \), we use the initial conditions \( y(0) = -3 \) and \( y'(0) = -\frac{43}{3} \).
- Substituting \( t = 0
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