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

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MAT244H1F1: INTRODUCTION TO ORDINARY DIFFERENTIAL

EQUATIONS

SECOND MIDTERM SOLUTIONS

Problem 1. Solve the initial value problem

y00 −y0−6y= 6

y(0) = 1,

y0(0) = 1.

Solution: The homogeneous linear equation is y00 −y0−6y= 0 which has

characteristic equation r2−r−6 = 0. We can factor this as (r−3)(r+ 2) = 0, and

so we ﬁnd that two generating solutions are y(t) = e3tand y(t) = e−2t. Hence all

solutions to this equation are of the form

y(t) = Ae3t+Be−2t,

for some constants Aand B. Moreover, it is clear that a particular solution to

y00 −y0−6y= 6 is the constant solution y(t) = −1, from where we deduce all

solutions to this equation are of the form

y(t) = Ae3t+Be−2t−1.

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

y0(t)=3Ae3t−2Be−2t.

Using the initial conditions we obtain a linear system in Aand B:

A+B−1=1,

3A−2B= 1,

which has a unique solution A= 1 and B= 1. In conclusion, the solution we look

for is

y(t) = e3t+e−2t−1.

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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