Question

Assume, y = y(x), the general solution of: 2y'' - 26y' + 84y = 0, is: (a) y = c_1e^x + c_2xe^{-4x} (b) y = c_1e^{8x} + c_2e^{9x} (c) y = c_1e^{4x} + c_2e^{5x} (d) y = c_1e^{6x} + c_2e^{7x} (e) y = c_1e^{2x} + c_2e^{3x} (f) y = c_1 cos(x) + c_2 sin(4x) (g) y = c_1 cos(4x) + c_2 sin(x)

          Assume, y = y(x), the general solution of:
2y'' - 26y' + 84y = 0, is:
(a) y = c_1e^x + c_2xe^{-4x}
(b) y = c_1e^{8x} + c_2e^{9x}
(c) y = c_1e^{4x} + c_2e^{5x}
(d) y = c_1e^{6x} + c_2e^{7x}
(e) y = c_1e^{2x} + c_2e^{3x}
(f) y = c_1 cos(x) + c_2 sin(4x)
(g) y = c_1 cos(4x) + c_2 sin(x)

Assume, y = y(x), the general solution of:
2y” - 26y' + 84y = 0, is:
(a) y = c1e^x + c2xe^-4x
(b) y = c1e^8x + c2e^9x
(c) y = c1e^4x + c2e^5x
(d) y = c1e^6x + c2e^7x
(e) y = c1e^2x + c2e^3x
(f) y = c1 cos(x) + c2 sin(4x)
(g) y = c1 cos(4x) + c2 sin(x)

Added by Aitor L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Khushbu Rani

Step 1

First, we rewrite the given differential equation in a more standard form: $$2\frac{d^2y}{dx^2} - 26\frac{dy}{dx} + 84y = 0$$ Show more…

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Assume, y = y(x), the general solution of 2y'' - 26y' + 84y = 0, is: (a) y = c_1e^x + c_2xe^{-4x} (b) y = c_1e^{8x} + c_2e^{9x} (c) y = c_1e^{4x} + c_2e^{5x} (d) y = c_1e^{6x} + c_2e^{7x} (e) y = c_1e^{2x} + c_2e^{3x} (f) y = c_1 cos(x) + c_2 sin(4x) (g) y = c_1 cos(4x) + c_2 sin(x)