y'' - 11y'' + 30y' = 60e^x, \newline y(0) = 30, y'(0) = 25, y''(0) = 28.
Added by Andrew B.
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We can do this by assuming a solution of the form y = e^(rx), where r is a constant. Plugging this into the equation, we get: r^2e^(rx) - 11re^(rx) + 30e^(rx) = 0 Factoring out e^(rx), we get: e^(rx)(r^2 - 11r + 30) = 0 This equation will be satisfied if Show more…
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