5. A mass weighing 16 pounds stretches a spring 3 inches. The medium offers a damping force that is numerically equal to 2 times the instantaneous velocity. The mass is released from equilibrium with a downward velocity of 3 inches per second. a. Determine the equation of motion. b. Determine the first time (after t = 0) when the mass first passes through equilibrium. 6. Solve the IVP. y" -8y +15y9te² = y(0) = -1, y'(0) = 3 7. Find the general solution of the ODE. x²y"-2y=3x² - 1 x > 0 8. Find the general solution of the ODE. y"+2y24y=12x+9-e4x 9. Solve the IVP. y"" +12y" +36y' = 0 y(0) = 0, y'(0) = 1, y"(0) = = -7 1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question
5.
A mass weighing 16 pounds stretches a spring 3 inches. The medium offers a
damping force that is numerically equal to 2 times the instantaneous velocity. The
mass is released from equilibrium with a downward velocity of 3 inches per
second.
a.
Determine the equation of motion.
b. Determine the first time (after t = 0) when the mass first passes through
equilibrium.
6.
Solve the IVP.
y" -8y +15y9te²
=
y(0)
=
-1, y'(0) = 3
7. Find the general solution of the ODE.
x²y"-2y=3x² - 1 x > 0
8.
Find the general solution of the ODE.
y"+2y24y=12x+9-e4x
9.
Solve the IVP.
y"" +12y" +36y' = 0
y(0) = 0, y'(0) = 1, y"(0) = = -7
1
Transcribed Image Text:5. A mass weighing 16 pounds stretches a spring 3 inches. The medium offers a damping force that is numerically equal to 2 times the instantaneous velocity. The mass is released from equilibrium with a downward velocity of 3 inches per second. a. Determine the equation of motion. b. Determine the first time (after t = 0) when the mass first passes through equilibrium. 6. Solve the IVP. y" -8y +15y9te² = y(0) = -1, y'(0) = 3 7. Find the general solution of the ODE. x²y"-2y=3x² - 1 x > 0 8. Find the general solution of the ODE. y"+2y24y=12x+9-e4x 9. Solve the IVP. y"" +12y" +36y' = 0 y(0) = 0, y'(0) = 1, y"(0) = = -7 1
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