MAT244H1 Study Guide - Midterm Guide: System Of Linear Equations, Wronskian, Scilab
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MAT244H1 Full Course Notes
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Solve the initial value problem y(cid:48)(cid:48) y(cid:48) 6y = 6 y(0) = 1, y(cid:48)(0) = 1. Solution: the homogeneous linear equation is y(cid:48)(cid:48) y(cid:48) 6y = 0 which has characteristic equation r2 r 6 = 0. We can factor this as (r 3)(r + 2) = 0, and so we nd that two generating solutions are y(t) = e3t and y(t) = e 2t. Hence all solutions to this equation are of the form y(t) = ae3t + be 2t, for some constants a and b. Notice that the constant solution y(t) = 1 can be obtained by the method of undetermined coe cients since 0 is not a root of the characteristic polynomial and the right hand side is a constant. Di erentiating the general solution we nd y(cid:48)(t) = 3ae3t 2be 2t. Using the initial conditions we obtain a linear system in a and b:
