y" – 10y' + 25y = 0, y(0) = - 4, - 4, y'(0) = - 15 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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![## Differential Equation Problem
We are given the following second-order linear differential equation with constant coefficients:
\[ y'' - 10y' + 25y = 0 \]
along with the initial conditions:
\[ y(0) = -4 \]
\[ y'(0) = -15 \]
We need to determine the function \( y(t) \).
**Solution:**
To solve this differential equation, follow these steps:
1. **Find the characteristic equation:**
The given differential equation is a homogeneous, linear differential equation with constant coefficients. The characteristic equation corresponding to this differential equation is:
\[ r^2 - 10r + 25 = 0 \]
2. **Solve the characteristic equation:**
Solve for the roots \( r \):
\[ (r - 5)^2 = 0 \]
\[ r = 5 \quad \text{(double root)} \]
3. **Form the general solution:**
Since we have a double root, the general solution will be of the form:
\[ y(t) = (C_1 + C_2t) e^{rt} \]
Substituting \( r = 5 \):
\[ y(t) = (C_1 + C_2t) e^{5t} \]
4. **Apply the initial conditions to determine constants:**
- Use \( y(0) = -4 \):
\[ y(0) = C_1 e^{0} + C_2(0)e^{0} = C_1 = -4 \]
- Use \( y'(0) = -15 \):
First, find \( y'(t) \):
\[ y'(t) = 5(C_1 + C_2t) e^{5t} + C_2 e^{5t} \]
\[ y'(t) = (5C_1 + 5C_2t + C_2) e^{5t} \]
\[ y'(t) = (5C_1 + C_2 + 5C_2t) e^{5t} \]
Substitute \( t = 0 \):
\[ y'(0) = (5C_1 + C_2) e^{0} \]
\[ y'(0) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62b092f3-63db-4d93-982f-67d0473d8e68%2Fb14cc422-5b43-4e4d-9d6e-c0fb7415742a%2Fdk9iplf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Differential Equation Problem
We are given the following second-order linear differential equation with constant coefficients:
\[ y'' - 10y' + 25y = 0 \]
along with the initial conditions:
\[ y(0) = -4 \]
\[ y'(0) = -15 \]
We need to determine the function \( y(t) \).
**Solution:**
To solve this differential equation, follow these steps:
1. **Find the characteristic equation:**
The given differential equation is a homogeneous, linear differential equation with constant coefficients. The characteristic equation corresponding to this differential equation is:
\[ r^2 - 10r + 25 = 0 \]
2. **Solve the characteristic equation:**
Solve for the roots \( r \):
\[ (r - 5)^2 = 0 \]
\[ r = 5 \quad \text{(double root)} \]
3. **Form the general solution:**
Since we have a double root, the general solution will be of the form:
\[ y(t) = (C_1 + C_2t) e^{rt} \]
Substituting \( r = 5 \):
\[ y(t) = (C_1 + C_2t) e^{5t} \]
4. **Apply the initial conditions to determine constants:**
- Use \( y(0) = -4 \):
\[ y(0) = C_1 e^{0} + C_2(0)e^{0} = C_1 = -4 \]
- Use \( y'(0) = -15 \):
First, find \( y'(t) \):
\[ y'(t) = 5(C_1 + C_2t) e^{5t} + C_2 e^{5t} \]
\[ y'(t) = (5C_1 + 5C_2t + C_2) e^{5t} \]
\[ y'(t) = (5C_1 + C_2 + 5C_2t) e^{5t} \]
Substitute \( t = 0 \):
\[ y'(0) = (5C_1 + C_2) e^{0} \]
\[ y'(0) =
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