Solve the following differential equation with initial conditions: y"e-2t+10e4t; y(0) = 1, y'(0) = 0. The solution is: y=e4t+e-21-3t+ A. B. y=e-2t + e4f-1 C. y=-2t+4-2t+ - -He-2t + het-2 D. E. =4e-2t +16e4t-6t+19
Solve the following differential equation with initial conditions: y"e-2t+10e4t; y(0) = 1, y'(0) = 0. The solution is: y=e4t+e-21-3t+ A. B. y=e-2t + e4f-1 C. y=-2t+4-2t+ - -He-2t + het-2 D. E. =4e-2t +16e4t-6t+19
Calculus: Early Transcendentals
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Chapter1: Functions And Models
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![### Differential Equations Problem
**Problem Statement:**
Solve the following differential equation with initial conditions:
\[ y'' = e^{-2t} + 10e^{4t}, \; y(0) = 1, \; y'(0) = 0. \]
The solution is:
**Options:**
- **A.** \[ y = \frac{5}{8}e^{4t} + \frac{1}{4}e^{-2t} - 3t + \frac{1}{8} \]
- **B.** \[ y = e^{-2t} + e^{4t} - 1 \]
- **C.** \[ y = \frac{1}{4}e^{-2t} + \frac{5}{8} e^{4t} - 2t + \frac{1}{8} \]
- **D.** \[ y = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2 \]
- **E.** \[ y = 4e^{-2t} + 16e^{4t} - 6t + 19 \]
**Explanation:**
To solve this differential equation, one must use techniques related to solving second-order differential equations with constant coefficients and particular solutions for non-homogeneous differential equations. The provided options include potential solutions that satisfy both the differential equation and the given initial conditions.
Students are encouraged to:
1. Verify the general solution of the associated homogeneous equation.
2. Find the particular solution that fits the non-homogeneous part.
3. Apply the initial conditions to determine the constants in the solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bf6a656-c902-430b-8919-3920c6e4552b%2F986c885e-6ed1-4ee4-81a4-28de5ff3bbcb%2Fht0nnlu_processed.png&w=3840&q=75)
Transcribed Image Text:### Differential Equations Problem
**Problem Statement:**
Solve the following differential equation with initial conditions:
\[ y'' = e^{-2t} + 10e^{4t}, \; y(0) = 1, \; y'(0) = 0. \]
The solution is:
**Options:**
- **A.** \[ y = \frac{5}{8}e^{4t} + \frac{1}{4}e^{-2t} - 3t + \frac{1}{8} \]
- **B.** \[ y = e^{-2t} + e^{4t} - 1 \]
- **C.** \[ y = \frac{1}{4}e^{-2t} + \frac{5}{8} e^{4t} - 2t + \frac{1}{8} \]
- **D.** \[ y = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2 \]
- **E.** \[ y = 4e^{-2t} + 16e^{4t} - 6t + 19 \]
**Explanation:**
To solve this differential equation, one must use techniques related to solving second-order differential equations with constant coefficients and particular solutions for non-homogeneous differential equations. The provided options include potential solutions that satisfy both the differential equation and the given initial conditions.
Students are encouraged to:
1. Verify the general solution of the associated homogeneous equation.
2. Find the particular solution that fits the non-homogeneous part.
3. Apply the initial conditions to determine the constants in the solution.
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