APPLICATION WORKSHEET: RESONANCE

110.302 DIFFERENTIAL EQUATIONS PROFESSOR RICHARD BROWN

Problem. Solve the Ordinary Differential Equation

80ẍ + 10000x = 2500 cos (\frac{πνt}{6}) .

Strategy. Solving the ODE means finding the general solution (the 1-parameter family of solutions). We place this linear, first order ODE into its standard form and compute the integrating factor. Then we multiply the ODE by its integrating factor and then integrate to find the general solution.

Solution. First, notice that the differential equation

(1)  80ẍ + 10000x = 2500 cos (\frac{πνt}{6})

is a second-order linear nonhomogeneous ODE with constant coefficients. To find the general solution, we will first find a set of two linearly independent solutions to the homogeneous version

80ẍ + 10000x = 0.

The characteristic equation for this ODE is then 80r^{2} + 10000 = 0, which is solved by r = ±\sqrt{−125} = ±5\sqrt{5} i. Being purely imaginary (and hence complex), we can immediately write out the general solution of this homogeneous, linear constant coefficient second order ODE, and get

x(t) = c_{1} cos 5\sqrt{5}t + c_{2} sin 5\sqrt{5}t.

The general solution to the original nonhomogeneous ODE is then

x(t) = c_{1} cos 5\sqrt{5}t + c_{2} sin 5\sqrt{5}t + X(t),

where X(t) is any solution to Equation ??. To find a suitable representative for X(t), we can use the Method of Constant Coefficients:

Choose X(t) = A cos (\frac{πνt}{6}). We would normally also include another constant involving the sine function, but since there is no first derivative in the ODE, there will not be a sine component (Stop here for a minute to absorb and accept this idea).

Now since Ẍ(t) = −A (\frac{πν}{6} )^{2} cos (\frac{πνt}{6}), we can sub these into the ODE to get

80 (−A (\frac{πν}{6} )^{2} cos (\frac{πνt}{6})) + 10000 (A cos (\frac{πνt}{6})) = 2500 cos (\frac{πνt}{6})
(10000 − 80 (\frac{πν}{6} )^{2}) A cos (\frac{πνt}{6}) = 2500 cos (\frac{πνt}{6})
A cos (\frac{πνt}{6}) = \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} cos (\frac{πνt}{6})

A = \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} .

Hence the general solution to Equation ?? is

x(t) = c_{1} cos 5\sqrt{5}t + c_{2} sin 5\sqrt{5}t + \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} cos (\frac{πνt}{6}) .

You can see how the speed of the car would affect the solution. In fact, there is a special value of ν for which this solution is NOT defined. That is, when 10000 − 80 (\frac{πν}{6} )^{2} = 0. This is solved by

10000 = 80 (\frac{πν}{6} )^{2}
\sqrt{\frac{10000}{80}} = \frac{πν}{6}
\frac{6\sqrt{125}}{π} = ν ≅ 21.35 feet per second ≅ 14.55 miles per hour.

For part (c), first we find the particular solution corresponding to the initial conditions x(0) = 0 and ẋ(0) = 0. For expediency, let’s start with the second one. We calculate:

ẋ(0) = \frac{d}{dt} [c_{1} cos 5\sqrt{5}t + c_{2} sin 5\sqrt{5}t + \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} cos (\frac{πνt}{6})] |_{t=0}

= (−5\sqrt{5}c_{1} sin 5\sqrt{5}t + 5\sqrt{5}c_{2} cos 5\sqrt{5}t − (\frac{πν}{6} ) \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} sin (\frac{πνt}{6})) |_{t=0}

= 5\sqrt{5}c_{2} cos 5\sqrt{5}(0) = 0

which implies that c_{2} = 0. Thus the solution is now

x(t) = c_{1} cos 5\sqrt{5}t + \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} cos (\frac{πνt}{6}) .

APPLICATION WORKSHEET: RESONANCE 3

As for the other initial condition, we get

x(0) = c_{1} cos 5\sqrt{5}(0) + \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} cos (\frac{πν(0)}{6})

= c_{1} + \frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} = 0

⟹ c_{1} = \frac{−2500}{10000 − 80 (\frac{πν}{6} )^{2}}

Thus our particular solution is

x(t) = (\frac{−2500}{10000 − 80 (\frac{πν}{6} )^{2}} ) cos 5\sqrt{5}t + (\frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} ) cos (\frac{πνt}{6})

= (\frac{2500}{10000 − 80 (\frac{πν}{6} )^{2}} ) (cos (\frac{πνt}{6}) − cos 5\sqrt{5}t) .