MATH 223 Lecture Notes - Lecture 3: Angular Frequency, Simple Harmonic Motion, Damping Ratio

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ECE 205 - Winter 2017 Assignment 3 Solutions Due 05-02-2017
1: Initial Value Problems
Determine the general solution of the following IVPs. You need only submit the solution. (10
points).
a)y′′(x) + 2y(x)15y(x) = x. y(0) = 2/225, y(0) = 0.
b)y′′(x) + 2y(x)15y(x) = x. y(1) = 2/225, y(1) = 1/15.
c)y′′(x) + 2y(x)15y(x) = x. y(0) = 2/225, y(1) = 2/225.
d)y′′(x) + 2y(x)15y(x) = x. y(0) = 2/225, y(1) = 0.
If we consider the homogeneous ODE, we have y′′ + 2y15y= 0. The characteristic equation is
m2+ 2m15 = (m+ 5)(m3) = 0. Therefore, the characteristic roots are m=5 and m= 3.
The general solution to the homogeneous ODE is thus:
yH(x) = Ae5x+Be3x
The right hand side of the original ODE is f(x) = x. Thus the particular solution is of the form
yP(x)α+βx. Differentiating yP(x) and substituting back into the ODE, we get β=1
15 and
α=2
225 . Thus the general solution is
y(x) = Ae5x+Be3x2
225 1
15x.
a) for the initial conditions y(0) = 2
225 and y(0) = 0, we get the following system of
equations in Aand B:
(A+B2
225 =2
225
5A+ 3B1
15 = 0
which yields A=1
120 and B=1
120 . The solution is therefore
y(x) = 1
120e5x+1
120e3x2
225 1
15x.
b) for the initial conditions y(1) = 2
225 and y(1) = 1
15 , we get the following system of
equations in Aand B:
(Ae5+Be32
225 1
15 =2
225
5Ae5+ 3Be31
15 =1
15
which yields A=1
40 e5and B=1
24 e3. The solution is therefore
y(x) = 1
40e55x+1
24e3+3x2
225 1
15x.
c) for the initial conditions y(0) = 2
225 and y(1) = 2
225 , we get the following system of
equations in Aand B:
(A+B2
225 =2
225
Ae5+Be32
225 1
15 =2
225
1
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ECE 205 - Winter 2017 Assignment 3 Solutions Due 05-02-2017
which yields A=e5
15(e81) and B=e5
15(e81) . The solution is therefore
y(x) = e5
15(e81)(e5x+e3x)2
225 1
15x.
d) for the initial conditions y(0) = 2
225 and y(1) = 0, we get the following system of
equations in Aand B:
(A+B2
225 =2
225
5Ae5+ 3Be31
15 = 0
which yields A=1
15 (e5
5+3e8) and B=1
15 (e5
5+3e8). The solution is therefore
y(x) = e5
15(5 + 3e8)(e5xe3x)2
225 1
15x.
2: Circuits
At time t= 0, a D.C. voltage of Vis applied to an LR circuit. Find the time taken for the
current to reach 95% of its final value. Submit the answer only.
The differential equation satisfying this problem is
LdI
dt +IR =V.
This is a first order linear ordinary differential equation with integrating factor µ(t) = eR
Lt.
Multiplying the equation (in standard form) by the integrating factor and integrating and
rearranging, gives the solution
I(t) = V
R+CeRt
L.
Since I(0) = 0, then C=V
R. The solution to the IVP is therefore
I(t) = V
R1eRt
L
. The final value is obtained as tapproaches infinity. Therefore, the final value is V
R. We are
looking for tsuch that I(t) = 0.95V
R. This gives
t=L
Rln(20).
3: Mass-Spring system
Consider a mass-spring system. The displacement of the mass from equilibrium is modelled by a
function x(t). Assuming that the mass is 1 unit, the spring constant is k > 0, the damping
coefficient is 2dwhere d > 0, then x(t) satisfies the ODE:
¨x+ 2d˙x+kx = 0,
2
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