Find the solution of the differential equation y'' + 6y' + 9y = 0 satisfying the initial conditions y(0) = 5, y'(0) = -18. Answer: y(t) = 5*e^(3*t)-33*t*e^(3*t) Your answer should be a function of t
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To do this, we assume that the solution has the form y(t) = e^(rt), where r is a constant. Substituting this into the differential equation, we get: r e^(rt) + 6 e^(rt) + 9 e^(rt) = 0 Factorizing e^(rt), we get: e^(rt) (r + 6 + 9) = 0 Since e^(rt) is never Show more…
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