Solve the following differential equation y''' - y'' + 2y' + 4y = 0 Select one: a. y = c1e^-3x + e^3x[c2 cos(3x) + c3 sin(3x)] b. y = c1e^-x + e^x[c2 cos(?3x) + c3 sin(?3x)] c. y = c1e^x + e^x[c2 cos(2x) + c3 sin(2x)] d. y = c1e^-3x + e^1.5x[c2 cos(2.6x) + c3 cos(2.6x)]
Added by Purificaci-N M.
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To solve it, we assume a solution of the form: y(x) = e^{rx} Taking the first and second derivatives, we get: y'(x) = re^{rx} y''(x) = r^2e^{rx} Now, we substitute these expressions back into the original differential equation: (r^2 + 2r + 4)e^{rx} = 0 Since Show more…
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