# 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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