d the general solution of this ODE: d?y dy + 13 + 42y = 4t2 + 1+ 2e dt? dt - 8t solution will be of the form: y(t) = Cy1(t) + Dy2(t) + Yp(t) use C and D as the arbitrary constants. y(t)
d the general solution of this ODE: d?y dy + 13 + 42y = 4t2 + 1+ 2e dt? dt - 8t solution will be of the form: y(t) = Cy1(t) + Dy2(t) + Yp(t) use C and D as the arbitrary constants. 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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Question
![**Find the general solution of this ODE:**
\[ \frac{d^2y}{dt^2} + 13 \frac{dy}{dt} + 42y = 4t^2 + 1 + 2e^{-8t} \]
The solution will be of the form:
\[ y(t) = C y_1(t) + D y_2(t) + y_p(t) \]
so use \( C \) and \( D \) as the arbitrary constants.
\[ y(t) = \]
**Explanation:**
This problem involves finding the general solution to a second-order linear ordinary differential equation (ODE) with non-homogeneous terms on the right-hand side. The given ODE is:
\[ \frac{d^2y}{dt^2} + 13 \frac{dy}{dt} + 42y = 4t^2 + 1 + 2e^{-8t} \]
1. **Homogeneous Part**:
- Solve the characteristic equation corresponding to the left-hand side of the ODE:
\[ \frac{d^2y}{dt^2} + 13 \frac{dy}{dt} + 42y = 0 \]
- Let the roots be \( r_1 \) and \( r_2 \).
2. **Particular Solution**:
- Find a particular solution \( y_p(t) \) to the non-homogeneous ODE.
3. **General Solution**:
- Construct the general solution by combining the solutions to the homogeneous and particular parts of the equation:
\[ y(t) = C y_1(t) + D y_2(t) + y_p(t) \]
- Here, \( y_1(t) \) and \( y_2(t) \) are the solutions to the homogeneous equation, and \( y_p(t) \) is the particular solution to the non-homogeneous equation.
- \( C \) and \( D \) are arbitrary constants determined by initial conditions.
Enter the final expression for \( y(t) \) in the provided box.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc700bc18-61e4-4e04-a31d-52f10b21c2d0%2F0ad815d4-6c95-4032-8424-782da637f287%2Fx3guozj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find the general solution of this ODE:**
\[ \frac{d^2y}{dt^2} + 13 \frac{dy}{dt} + 42y = 4t^2 + 1 + 2e^{-8t} \]
The solution will be of the form:
\[ y(t) = C y_1(t) + D y_2(t) + y_p(t) \]
so use \( C \) and \( D \) as the arbitrary constants.
\[ y(t) = \]
**Explanation:**
This problem involves finding the general solution to a second-order linear ordinary differential equation (ODE) with non-homogeneous terms on the right-hand side. The given ODE is:
\[ \frac{d^2y}{dt^2} + 13 \frac{dy}{dt} + 42y = 4t^2 + 1 + 2e^{-8t} \]
1. **Homogeneous Part**:
- Solve the characteristic equation corresponding to the left-hand side of the ODE:
\[ \frac{d^2y}{dt^2} + 13 \frac{dy}{dt} + 42y = 0 \]
- Let the roots be \( r_1 \) and \( r_2 \).
2. **Particular Solution**:
- Find a particular solution \( y_p(t) \) to the non-homogeneous ODE.
3. **General Solution**:
- Construct the general solution by combining the solutions to the homogeneous and particular parts of the equation:
\[ y(t) = C y_1(t) + D y_2(t) + y_p(t) \]
- Here, \( y_1(t) \) and \( y_2(t) \) are the solutions to the homogeneous equation, and \( y_p(t) \) is the particular solution to the non-homogeneous equation.
- \( C \) and \( D \) are arbitrary constants determined by initial conditions.
Enter the final expression for \( y(t) \) in the provided box.
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