Question

If the constant-coefficient, linear differential equation y^(5) + y^(p1) - 2y^(p2) + ky^(p3) - 8y^(p4) = 0 has the general solution y = (A + Bx)e^(-x) + Ce^(2x) + D cos 2x + E sin 2x, find the values of k, p3 and p4. (a) k = -12, p3 = 1, p4 = 0 (b) k = -12, p3 = 2, p4 = 0 (c) k = 12, p3 = 2, p4 = 1 (d) k = -12, p3 = 1, p4 = 1 (e) k = 12, p3 = 1, p4 = 0

          If the constant-coefficient, linear differential equation
y^(5) + y^(p1) - 2y^(p2) + ky^(p3) - 8y^(p4) = 0
has the general solution
y = (A + Bx)e^(-x) + Ce^(2x) + D cos 2x + E sin 2x,
find the values of k, p3 and p4.
(a) k = -12, p3 = 1, p4 = 0
(b) k = -12, p3 = 2, p4 = 0
(c) k = 12, p3 = 2, p4 = 1
(d) k = -12, p3 = 1, p4 = 1
(e) k = 12, p3 = 1, p4 = 0

If the constant-coefficient, linear differential equation
y^(5) + y^(p1) - 2y^(p2) + ky^(p3) - 8y^(p4) = 0
has the general solution
y = (A + Bx)e^(-x) + Ce^(2x) + D cos 2x + E sin 2x,
find the values of k, p3 and p4.
(a) k = -12, p3 = 1, p4 = 0
(b) k = -12, p3 = 2, p4 = 0
(c) k = 12, p3 = 2, p4 = 1
(d) k = -12, p3 = 1, p4 = 1
(e) k = 12, p3 = 1, p4 = 0

Added by Kristine J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Bradley Duda

Step 1

First, let's rewrite the given differential equation in a more standard form: y^5 + yp^1 - 2yp^2 + kyp^3 - 8yp^4 = 0 Show more…

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If the constant-coefficient linear differential equation y^5 + yp^1 - 2yp^2 + kyp^3 - 8yp^4 = 0 has the general solution xuIS + xsoa + x + x - + v = f find the values of kp^3 and p^4 (a) (b) (c) (d) (e) k = -12, P^3 = 1, p^4 = 0 k = 12, P^3 = 2, p^4 = 0 k = 12, P^3 = 2, P^4 = 1 k = -12, P^3 = 1, p^4 = 1 k = 12, P^3 = 1, p^4 = 0