MATH 223 Lecture Notes - Lecture 3: Angular Frequency, Simple Harmonic Motion, Damping Ratio
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21 Oct 2020
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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′′ + 2y′−15y= 0. The characteristic equation is
m2+ 2m−15 = (m+ 5)(m−3) = 0. Therefore, the characteristic roots are m=−5 and m= 3.
The general solution to the homogeneous ODE is thus:
yH(x) = Ae−5x+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) = Ae−5x+Be3x−2
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+B−2
225 =−2
225
−5A+ 3B−1
15 = 0
which yields A=−1
120 and B=1
120 . The solution is therefore
y(x) = −1
120e−5x+1
120e3x−2
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:
(Ae−5+Be3−2
225 −1
15 =−2
225
−5Ae−5+ 3Be3−1
15 =−1
15
which yields A=1
40 e5and B=1
24 e−3. The solution is therefore
y(x) = 1
40e5−5x+1
24e−3+3x−2
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+B−2
225 =−2
225
Ae−5+Be3−2
225 −1
15 =−2
225
1
ECE 205 - Winter 2017 Assignment 3 Solutions Due 05-02-2017
which yields A=−e5
15(e8−1) and B=e5
15(e8−1) . The solution is therefore
y(x) = e5
15(e8−1)(−e−5x+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+B−2
225 =−2
225
−5Ae−5+ 3Be3−1
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)(e−5x−e3x)−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+Ce−Rt
L.
Since I(0) = 0, then C=−V
R. The solution to the IVP is therefore
I(t) = V
R1−e−Rt
L
. The final value is obtained as tapproaches infinity. Therefore, the final value is V
R. We are
looking for t⋆such 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,
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