Question

Find the solution of the given initial value problem: y''' + y' = sec(t), y(0) = 6, y'(0) = 3, y''(0) = -4. y(t) = 2 + 4 \cos(t) + 4 \sin(t) - t \cos(t) + \sin(t) \ln(\cos(t))

          Find the solution of the given initial value problem:
y''' + y' = sec(t), y(0) = 6, y'(0) = 3, y''(0) = -4.
y(t) = 2 + 4 \cos(t) + 4 \sin(t) - t \cos(t) + \sin(t) \ln(\cos(t))

Added by Olivia L.

Elementary Differential Equations and Boundary Value Problem

William E. Boyce, Richard C. DiPrima 7th Edition

Chapter 2

Instant Answer

Solved by Expert Kelan H

Step 1

The characteristic equation is r^2 + r = 0. Factoring out an r, we get r(r + 1) = 0. So the solutions are r = 0 and r = -1. Therefore, the general solution of the homogeneous equation is y_h(t) = c1e^(-t) + c2, where c1 and c2 are constants. Show more…

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Find the solution of the given initial value problem: y" + y' = sec(t), y(0) = 6, y'(0) = 3, y"(0) = -4. y(t) = 2 + 4cos(t) + 4sin(t) - tcos(t) + sin(t)ln(cos(t))