MATH 1106 DIFFERENTIAL EQUATIONS NOTES CONSTANT COEFFICIENT SECOND ORDER LINEAR

Contents 1. 2. 3. 4. 5. 6. 7. 8.

Introduction Linearity Complex roots Where These Come Up Overdamped vs. Underdamped Linearity – the Inhomogeneous Case Inhomogeneous Equation and Resonance Second Order Equations vs. Systems of First Order

1 2 3 4 5 7 8 8

1. Introduction Differential equations like dx d2 x + 3x = 0 +4 dt2 dt suggest what we want to work with. It turns out that these kinds of equations are (1) easy to solve (2) show up widely in applications In a later section we’ll say more about where they arise. The equation is called second order because derivatives of second order appear (but no higher.) Constant coefficient because the factors in front of the derivatives do not depend on t. Linear, like in the first order case refers precisely to the form of the equation; more on that in section 2. It’s easy to find some solutions of an equation like the one above. Seek a solution of the form x = ert . Since x=

we see that x = ert

ert

dx = rert dt d2 x = r2 ert dt2 is a solution exactly when r2 ert + 4rert + 3ert = 0.

The ert factors out and we have a solution exactly when r satisfies the characteristic equation r2 + 4r + 3 = 0. 1

2

CONSTANT COEFFICIENT SECOND ORDER LINEAR

Using the factorization r2 + 4r + 3 = (r + 3)(r + 1), the solutions are r = −1 and r = −3. Thus x = e−t and x = e−3t are both solutions to the differential equation. Note that for this kind of differential equation, no matter how the numbers change, the characteristic equation is always quadratic, so we can never get stuck. Worst case, the quadratic formula √ −b ± b2 − 4ac r= 2a will give us the solution to the quadratic equation ar2 + br + c = 0. Here are some exercises. View them as like “look at”’ problems. You should not hand them in unless they are assigned as part of supplementary problems F. Exercise A: (To Be Continued in Exercise B below.) Use this trial of ert and the resultant characteristic equation to find some solutions of the following differential equations: (1) (2) (3) (4) (5)

d2 x dt2 d2 x dt2 d2 x dt2 d2 x dt2 d2 x dt2

−x=0 − 4x = 0 − 4 dx dt + 3x = 0 + 4 dx dt + x = 0 dx + 4 dt = 0. 2. Linearity

In our first example, the solutions r = −1 and r = −3 to our characteristic equation gave rise to two different solutions to dx d2 x +4 + 3x = 0 2 dt dt Let’s give these solutions two different names x1 (t) = e−t and x2 (t) = e−3t . We can build more solutions out of them. For example since x2 (t) = e−3t satisfies d2 x2 dx2 +4 + 3x2 = 0, 2 dt dt if we multiply this equation by e.g. the constant 10, we see that x(t) = 10x2 (t) = 10e−3t also solves the differential equation. Even more like this is true. We can add the two equations together to see that x(t) = x1 (t) + x2 (t) is also a solution: d2 x1 dt2

+4

dx1 dt

+3x1

= 0

d2 x2 dt2

+4

dx2 dt

+3x2

= 0

+

shows

d2 (x1 + x2 ) d(x1 + x2 ) +4 +3(x1 + x2 ) = 0 2 dt dt The upshot of all this is that whenever we have two solutions x1 (t) and x2 (t), we can choose any two constants c1 and c2 and get a new solution x(t) = c1 x1 (t) + c2 x2 (t). So in our example, we obtain x(t) = c1 e−t + c2 e−3t .

MATH 1106 DIFFERENTIAL EQUATIONS NOTES

3

Here we really want x1 (t) and x2 (t) to be “independent” solutions like the ones we get from two different roots of the characteristic equation. A solution like this for a second order differential equation is “good” because it can be shown that the most general solution to a second order differential equation usually depends on two arbitrary constants. So we refer to x(t) = c1 e−t + c2 e−3t as the general solution of the differential equation dx2 d2 x2 +4 + 3x2 = 0. 2 dt dt As with first order equations, we could use initial values like the values of x(0) and x0 (0) to determine c1 and c2 , but we won’t work with that too much. Exercise B: For each of the five differential equations in Exercise A above, use this technique to produce the general solutions.

3. Complex roots As you know, quadratic equations sometimes have imaginary or complex roots. For example r2 + 1 = 2

r = r=

0 −1 ±i

√ where i = −1. It turns out solutions involving i0 s are very exciting - they give rise to oscillating (sin and cos) type solutions. The rule is that a pair of complex roots α ± ωi (α and ω assumed to be real numbers) give us a pair of solutions x1 (t) = eαt cos ωt

x2 (t) = eαt sin ωt

Example 1: Find the general solution of d2 x +x=0 dt2 Solution: As above, trying x(t) = ert leads to the characteristic equation r2 +1 = 0 with solutions r = ±i (which might also be thought of as 0 ± (1)i.) So our building block solutions are x1 (t) = e0t cos (1)t = cos t and x2 (t) = e0t sin (1)t = sin t leading to a general solution of x(t) = c1 cos t + c2 sin t. Example 2: Find the general solution of d2 x dx +2 + 5x = 0 dt2 dt Solution: As above, trying x(t) = ert leads to the characteristic equation r2 + 2r + 5 = 0. The quadratic formula gives us the solutions

4

CONSTANT COEFFICIENT SECOND ORDER LINEAR

r=

−2 ±

= = =

p

22 − 4(5) 2 √ −2 ± −16 2 −2 ± 4i 2 −1 ± 2i.

So our building block solutions are x1 (t) = e−t cos 2t and x2 (t) = e−t sin 2t leading to a general solution of x(t) = e−t (c1 cos 2t + c2 sin 2t). The idea of complex solutions α ± ωi being associated with solutions involving sines and cosines is suggested by Euler’s formula eiθ = cos θ + i sin θ which explains that sines and cosines are closely related to exponentials of imaginary numbers. Exercise C: Find the general solution of the following differential equations: (1) (2) (3) (4)

d2 x dt2 d2 x dt2 d2 x dt2 d2 x dt2

+ 4x = 0 + 3x = 0 − 2 dx dt + 5x = 0 + 4 dx dt + 13x = 0 4. Where These Come Up

One important case is the damped simple harmonic oscillator, such as that pictured below. We think of a mass m attached to a spring, with x measuring how far from the equilibrium position the mass is. As x increases in some direction, the spring supplies a restoring force −kx (proportional to the displacement) in the opposite direction. No damping so far and Newton’s Law gives d2 x = −kx. dt2 Motion of a Damped Harmonic Oscillator m

MATH 1106 DIFFERENTIAL EQUATIONS NOTES

5

In the picture, we imagine the mass might be immersed in some fluid, and so there is also a damping force (proportional to the velocity.) This leads to the damped harmonic oscillator dx d2 x m 2 +c + kx = 0. dt dt Even if you are not very interested in this mechanical situation, it turns out to be one of the most fundamental to think about for any system near an equilibrium. An electrical analogue appears in electric circuits like the one below that has capacitors (which can accumulate charge), resistors (which can damp out currents) and inductors (sensitive to changes in currents.) With a voltage source E(t), this circuit leads to the differential equation L

dx x d2 x +R + = E(t) 2 dt dt C

and x(t) is keeping track of the amount of charge accumulating on the capacitor, so that its derivative is the current flowing. Charges and Currents in a Circuit

Analyzing this kind of circuit is basic to understanding things like how a tuning circuit in a radio can select out just one frequency (the station you tuned to.) Another place where this kind of equation comes up is in the Glucose Tolerance Tests which have been used to diagnose diabetes. The equation which appears is dg d2 g + 2α + ω02 g = 0 2 dt dt where g(t) is keeping track of glucose level changes after a fast is broken. By studying the measured values of g(t) one finds out something about the constants in the model, which eventually leads to the diagnoses. You can read more about this if you are interested on pages 172-179 in the second edition of Martin Braun’s Differential Equations. 5. Overdamped vs. Underdamped Consider the damped oscillator equation m

d2 x dx +c + kx = 0. dt2 dt

6

CONSTANT COEFFICIENT SECOND ORDER LINEAR

The characteristic equation is mr2 + cr + k = 0 and the quadratic formula gives solutions

r=

−c ±

√

c2 − 4mk . 2m

The nature of these solutions will be very sensitive to whether c2 −4mk is negative or positive. The negative case is referred to as underdamped, and we have complex roots with oscillating (sin/cos) solutions. For example the general solution to

dx d2 x + 0.2 + 1.01x = 0 2 dt dt

is x(t) = e−.1t (c1 cos t + c2 sin t) and a typical underdamped solution might look like:

The overdamped case is illustrated by

d2 x dx +x=0 + 2.5 2 dt dt

with general solution x(t) = c1 e−2t + c2 e−.5t typically looking like:

MATH 1106 DIFFERENTIAL EQUATIONS NOTES

7

(The borderline case, called critical damping when c2 − 4mk = 0 has slightly more complicated solutions that we won’t go into.) 6. Linearity – the Inhomogeneous Case Almost all the equations we have dealt with thusfar had a right hand side of 0. The equation d2 x dx a 2 +b + cx = 0 dt dt is called the homogeneous equation. We can use the procedure we’ve talked about to get the general solution xh (t) to the homogeneous equation. For example, we showed earlier that xh = c1 cos t + c2 sin t was the general solution to d2 x + x = 0. dt2 Replacing the 0 on the right hand side by some function f (t) gives what is called an inhomogeneous equation dx d2 x + cx = f (t). +b dt2 dt In the mechanical case, one might refer to f (t) as the driving force. Again, linearity helps us. For example if xp (t) is one particular solution of the inhomogeneous equation and xh (t) is the general solution of the homogeneous, adding d2 xp dxp a 2 +b +cxp = f (t) dt dt + a

a

d2 xh dt2

+b

dxh dt

+cxh

= 0

8

CONSTANT COEFFICIENT SECOND ORDER LINEAR

shows

d(xp + xh ) d2 (xp + xh ) +b +c(xp + xh ) = f (t). dt2 dt In words, we are getting the general solution to the inhomogeneous equation by adding one particular solution xp to the general solution xh of the homogeneous. For example, by trying for a solution of the form x(t) = Ae2t to a

d2 x + x = e2t dt2 we see that it works if 4Ae2t + Ae2t = e2t which checks when 5A = 1, ie. A = 0.2. Thus xp (t) = 0.2e2t is a particular solution to the inhomogeneous, and using our previous general solution to the homogeneous, our general inhomogeneous solution is x(t) = 0.2e2t + c1 cos t + c2 sin t. 7. Inhomogeneous Equation and Resonance Let’s think about the damped harmonic oscillator with a driving force F0 cos ωt. It turns out that the character of the solutions here depend quite sensitively on the frequency ω. In many cases, a fairly small value of F0 at the right frequency can lead to large solutions. This is the phenomenon of resonance. Explicitly, using the ideas we have been talking about (and a fair bit of calculation), one can find a particular solution to m

d2 x dx +c + kx = F0 cos ωt dt2 dt

as x(t) = A cos (ωt − δ) where F0

A=

1

[(k − mω 2 )2 + c2 ω 2 ] 2 and the phase factor 

 c . k − mω 2 If, for example c is small, the amplitude A (as a function of ω) will get comparatively big when ω is close to a frequency where k −mω 2 = 0. This is what resonance is about. δ = arctan

8. Second Order Equations vs. Systems of First Order d2 x dv dx , then the acceleration 2 = If we give a “name” like v (for velocity) to dt dt dt and our second order equation a

d2 x dx +b + cx = 0 2 dt dt

may be re-expressed as a

dv + bv + cx = 0. dt

MATH 1106 DIFFERENTIAL EQUATIONS NOTES

9

In rearranged form, we are dealing with the system of first order differential equations dx = v dt dv = −cx −bv dt which is of the same general form as the predator prey systems dx dt dy dt

= f (x, y) = g(x, y)

we met in section 10.4. Second order differential equations are just another language for this kind of system.