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Find general solution of y" + 2y' + y = 2e^-t

Answer: c1e^-t + c2t^-t + t^2e^-t
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This is second order diff equatin which means if you break it down in the form of ay'' + by' + cy  you get general solutions of -b/2a or -1. (the b^2-4ac terms disappear from the quadratic formula). That means that by definition the general solution is C1e^-t + c2te^-t. The double integral of 2e^-t is t^2e^-t (first integral is 2te^-t).
Add those terms together for the total solution. Hope that helps!