MAT244H1 Study Guide - Midterm Guide: System Of Linear Equations, Wronskian, Scilab

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Published on 25 Oct 2018
Problem 1. Solve the initial value problem
y00 y06y= 6
y(0) = 1,
y0(0) = 1.
Solution: The homogeneous linear equation is y00 y06y= 0 which has
characteristic equation r2r6 = 0. We can factor this as (r3)(r+ 2) = 0, and
so we find that two generating solutions are y(t) = e3tand y(t) = e2t. Hence all
solutions to this equation are of the form
y(t) = Ae3t+Be2t,
for some constants Aand B. Moreover, it is clear that a particular solution to
y00 y06y= 6 is the constant solution y(t) = 1, from where we deduce all
solutions to this equation are of the form
y(t) = Ae3t+Be2t1.
Notice that the constant solution y(t) = 1 can be obtained by the method of
undetermined coefficients since 0 is not a root of the characteristic polynomial and
the right hand side is a constant.
Differentiating the general solution we find
Using the initial conditions we obtain a linear system in Aand B:
3A2B= 1,
which has a unique solution A= 1 and B= 1. In conclusion, the solution we look
for is
y(t) = e3t+e2t1.
Problem 2. (a): Find a solution of the equation
x2y00 x(x+ 2)y0+ (x+ 2)y= 0.
of the form y=ax +b, where aand bare constants.
(b): Find the general solution of the equation
x2y00 x(x+ 2)y0+ (x+ 2)y= 2x3e2x.
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