Published on 20 Nov 2012

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Quiz 3a Solutions

1. This question concerns second-order linear homogeneous equations. Show your

work.

(a) Find two linearly independent solutions of y′′ + 6y′+ 9y= 0.

The characteristic equation is r2+ 6r+ 9 = 0, which factors as (r+ 3)2= 0,

so there is a double root r=−3. Thus two linearly independent solutions

are given by

y1=e−3x, y2=xe−3x

(b) Find the general solution of y′′ + 4y= 0.The characteristic equation

is r2+ 4 = 0, with complex conjugate roots r=±2i. Thus two linearly

independent solutions are given by

y1= cos 2x , y2= sin 2x

and the general solution is

y=c1y1+c2y2

=c1cos 2x+c2sin 2x

2. The solution of x′′ + 4x= 0, with initial conditions x(0) = 1,x′(0) = −2, is

x(t) = cos ω0t−sin ω0t

where I haven’t told you what ω0is, but we can ﬁgure it out.

(a) What is ω0?

We might remember that the diﬀerential equation is in the form x′′ +ω2

0x= 0,

thus ω0= 2. Or we might observe that x′′ =−ω2

0cos ω0t+ω2

0sin ω0t, and

since this is a solution of x′′ + 4x= 0, we must have ω2

0= 4, thus ω0= 2.

(b) Find numbers C,αsuch that x(t)is written x(t) = Ccos(ω0t−α). (Hint:

we must have Ccos α= 1,Csin α=−1; drawing the unit circle may help

you.)

We must ﬁnd C,αsuch that Ccos α= 1, Csin α=−1. Thus C=

p

12+ (−1)2=

√

2. Since cos α > 0 and sin α < 0, the angle αis in

the fourth quadrant, with tan α= sin α/ cos α=−1, thus α= 7π/4. So

x(t) =

√

2 cos(2t−7π/4)

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