Solve the initial value problem.\\ $\frac{dy}{dt} + 2y = 5, y(0) = 1$\ The solution is $y = $
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This can be done by assuming a solution of the form y = e^(rt), where r is a constant. Substituting this into the differential equation gives: r e^(rt) + 2 e^(rt) = 0 Factor out e^(rt) to get: e^(rt) (r + 2) = 0 Since e^(rt) is never zero, we must have r + 2 = Show more…
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