Second-Order Equation Consider the second-order linear differential equation y″ − y' − 2y = 0.
(a) Verify that y = e2t is a solution; then check that y = e-t is a solution as well.
(b) Verify that y = Ae2t and y = e2t + e−1 are both solutions, where A is any real constant.
(c) Verify that for any constants A and B, a solution is y = Ae2t + Be−t.
(d) Determine values for A and B so that the solution of part (c) satisfies both y(0) = 2 and y′(0) = −5.
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