4 (20) Find the general solution to: y" 3y' 4y = e-*
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y" = r^2e^(rt) y' = re^(rt) Substituting these into the differential equation, we get: r^2e^(rt) + 3re^(rt) + 4e^(rt) = e^(-t) Dividing both sides by e^(rt), we get: r^2 + 3r + 4 = e^(-2t) This is the characteristic equation. We can solve for the roots Show more…
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