What is the solution of y'' - 4y' + 3y = 0, y(0) = 2, y'(0) = 3?
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Find the solution of the initial value problem: y'' - 4y' + 3y = 0, y(0) = 2, y'(0) = 3
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Solution Summary
This shows how to find the general solution for the equation.
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The characteristic equation is given as ar^2 + br + c = 0
or r^2 - 4r + 3 = 0
or (r-1)(r-3) = 0
Therefore, r1 = 1, r2 = ...
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