Solve the initial-value problem. y + 6y + 5y=0; y(0) = 0, y (0) = 3
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 1CR
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![### Initial-Value Problem
#### Problem Statement:
Solve the initial-value problem given by the differential equation and initial conditions:
\[
y'' + 6y' + 5y = 0; \quad y(0) = 0, \quad y'(0) = 3
\]
#### Explanation:
This is a second-order linear homogeneous differential equation. The general form is provided alongside specific initial conditions to solve for the particular solution:
1. \(y''\) denotes the second derivative of \(y\) (acceleration).
2. \(6y'\) denotes six times the first derivative of \(y\) (velocity).
3. \(5y\) denotes five times the function \(y\) itself (displacement).
The initial conditions are:
- \(y(0) = 0\): This means that when \(t = 0\), the value of \(y\) is \(0\).
- \(y'(0) = 3\): This means that when \(t = 0\), the value of the first derivative of \(y\), or the initial rate of change of \(y\), is \(3\).
### Steps to Solve:
1. **Find the characteristic equation**:
Convert the differential equation into a characteristic polynomial.
\[
r^2 + 6r + 5 = 0
\]
2. **Solve the characteristic equation**:
Find the roots \(r\) which satisfy the polynomial equation.
\[
r^2 + 6r + 5 = 0 \Rightarrow (r+1)(r+5) = 0 \Rightarrow r_1 = -1, \, r_2 = -5
\]
3. **Write the general solution**:
Using the roots of the characteristic equation, the general solution to the differential equation is:
\[
y(t) = C_1 e^{-t} + C_2 e^{-5t}
\]
4. **Apply the initial conditions**:
To find \(C_1\) and \(C_2\), use the given initial conditions.
- \(y(0) = 0\):
\[
y(0) = C_1 e^0 + C_2 e^0 = C_1 + C_2 = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa59c8a62-6d27-4c14-9dc5-c4241c4a7fe0%2Fe7e48e80-d370-47c7-9111-551151f12a3c%2Fb4pd60e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Initial-Value Problem
#### Problem Statement:
Solve the initial-value problem given by the differential equation and initial conditions:
\[
y'' + 6y' + 5y = 0; \quad y(0) = 0, \quad y'(0) = 3
\]
#### Explanation:
This is a second-order linear homogeneous differential equation. The general form is provided alongside specific initial conditions to solve for the particular solution:
1. \(y''\) denotes the second derivative of \(y\) (acceleration).
2. \(6y'\) denotes six times the first derivative of \(y\) (velocity).
3. \(5y\) denotes five times the function \(y\) itself (displacement).
The initial conditions are:
- \(y(0) = 0\): This means that when \(t = 0\), the value of \(y\) is \(0\).
- \(y'(0) = 3\): This means that when \(t = 0\), the value of the first derivative of \(y\), or the initial rate of change of \(y\), is \(3\).
### Steps to Solve:
1. **Find the characteristic equation**:
Convert the differential equation into a characteristic polynomial.
\[
r^2 + 6r + 5 = 0
\]
2. **Solve the characteristic equation**:
Find the roots \(r\) which satisfy the polynomial equation.
\[
r^2 + 6r + 5 = 0 \Rightarrow (r+1)(r+5) = 0 \Rightarrow r_1 = -1, \, r_2 = -5
\]
3. **Write the general solution**:
Using the roots of the characteristic equation, the general solution to the differential equation is:
\[
y(t) = C_1 e^{-t} + C_2 e^{-5t}
\]
4. **Apply the initial conditions**:
To find \(C_1\) and \(C_2\), use the given initial conditions.
- \(y(0) = 0\):
\[
y(0) = C_1 e^0 + C_2 e^0 = C_1 + C_2 = 0
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