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Example 5 - Mass-Spring System

Suppose a mass-spring system

mx" + bx' + kx

has spring constant k = 5, damping constant b = 4, and mass m = 1. At time t = 0, the position is x(0) = 1 and the velocity is x'(0) = 2. Find the position x(t) as a function of time.

The differential equation is

x" + 4x' + 5x = 0.

Step 1

The characteristic polynomial z2 + 4z + 5 has roots -2 - i and -2 + i. (These roots can be found using the quadratic equation.)

Step 2

The general solution is x(t) = e-2t[A cos t + B sin t].

Step 3

Find A and B using the given initial values. 1 = e0[A cos 0 + B sin 0], A = 1. Compute x'(t) and substitute to find B.

x'(t) = -2e-2t[-4 cos t + B sin t] + e-2t[-A sin t + B cos t], 2 = -2e0[cos 0 + B sin 0] + e0[-sin 0 + B cos 0], B = 4.

The particular solution is x(t) = e-2t[cos t + 4 sin t]. This is a damped oscillation with period 2π.

Let us now justify the solution method given at the beginning of this section. We may take the coefficient of y" to be one and consider the second order differential equation

(2)

y" + by' + cy = 0.

Let r and s be the roots of the characteristic polynomial, so that

z2 + bz + c = (z - r)(z - s) = z2 - rz - sz + rs.

If r and s are distinct (either two different real numbers or complex conjugates), we must show that the general solution is

y = Aert + Best.

If r = s, we must show that the general solution is

y = Aert + Btert.

The plan is to break the second order equation (2) into a pair of first order differential equations. It is useful to use the symbol D for the first derivative and D2 for the second derivative with respect to t. Thus

Dy = y', D2y = y".

The differential equation (2) can then be written in the form

(3)

(D2 +bD + c)y = 0.

We now wish to "factor" the expression D2 + bD + c as if it were a polynomial. It is to be understood that (D - r)(D - s)y means (D - r)u where u is the function (D - s)y = y' - sy. Thus, using the Sum and Product Rules for derivatives,

(D - r)(D - s)y = (y' - sy)' - r(y' - sy) = y" - sy' - ry' + rsy = y" + by' + cy = (D2 + bD + c)y.

This shows that the second order equation (3) is equivalent to the pair of first order equations

(4)

(D - r)u = 0,

(5)

(D - s)y = u.

Equation (4) is a homogeneous linear equation whose general solution is

w = Kert. Equation (5) may now be put in the form

(6)

y' - sy = Kert.

This first order linear equation was solved in Example 3 in Section 14.3. In the case r ≠ s, the general solution came out to be

14_differential_equations-204.gif

Putting A = K/(r - s), we get the general solution

y = Aert + Best.

In case r = s, the general solution is

y = Atest + Best,

where A = K. In each case we have the required formula for the general solution of the original equation (2).
Home Differential Equations Second Order Homogeneous Linear Equations Examples Example 5 - Mass-Spring System

Last Update: 2006-11-16