Mathematics 304 – Ordinary Differential Equations Problem Set 5 – Selected Solutions November 13, 2000

Section 4.8

56. a) We want to find the general solution of the ODE:

(1)  mx′′ + bx′ + kx = sin(βt)

The homogeneous equation mx′′ + bx′ + kx = 0 has auxiliary equation mr^{2} + br + k = 0, so

r = \frac{−b ± \sqrt{b^{2} − 4mk}}{2m} .

Since we are given b^{2} < 4mk, the two values for r are complex conjugates, and we can write the general solution of the homogeneous equation as

c_{1}e^{\frac{−bt}{2m}} cos(ωt) + c_{2}e^{\frac{−bt}{2m}} sin(ωt)

where ω = \frac{\sqrt{4mk−b^{2}}}{2m}.

To solve the inhomogeneous equation, we use the undetermined coefficients method. By form III in the table on page 208 of the text, we try a particular solution of the form

x_{p} = A cos(βt) + B sin(βt)

(Note that neither term here can solve the homogeneous equation – no exponential terms – hence we take s = 0 in the undetermined coefficients formula.) Substituting x_{p} into (1), we get:

\begin{aligned}(k − mβ^{2})A + bβB & = 0 \\ −bβA + (k − mβ^{2})B & = 1 \end{aligned}

The simplest way to write down the solution of a system like this is to use Cramer’s Rule:

\begin{aligned}A & = \begin{matrix}det & ( \begin{matrix}0 & bβ \\ 1 & k − mβ^{2} \end{matrix} ) \\ det & ( \begin{matrix}k − mβ^{2} & bβ \\ −bβ & k − mβ^{2} \end{matrix} ) \end{matrix} \\ & = \frac{−bβ}{(k − mβ^{2})^{2} + (bβ)^{2}} \end{aligned}

and

\begin{aligned}B & = \begin{matrix}det & ( \begin{matrix}k − mβ^{2} & 0 \\ −bβ & 1 \end{matrix} ) \\ det & ( \begin{matrix}k − mβ^{2} & bβ \\ −bβ & k − mβ^{2} \end{matrix} ) \end{matrix} \\ & = \frac{k − mβ^{2}}{(k − mβ^{2})^{2} + (bβ)^{2}} \end{aligned}

Then the general solution of the equation (1) is

c_{1}e^{\frac{−bt}{2m}} cos(ωt) + c_{2}e^{\frac{−bt}{2m}} sin(ωt) + A cos(βt) + B sin(βt),

where A, B are given by the formulas above.

b) Because b/2m > 0, the exponentials in the general solution tend to zero as t → ∞. The solution tends to A cos(βt) + B sin(βt). This is a sinusoidal function of the form C = cos(βt + ϕ) for suitable C, ϕ.

57. The idea here is to follow the same method used in problem 55. We first solve the IVP:

(2)  \{ \begin{matrix}y′′ + y = cos(V t) \\ y(−π/(2V)) = 0 \\ y′(−π/(2V)) = 0 \end{matrix}

(Note: it is given that y(t) = 0 for all t ≤ −π/(2V) – the car is not moving vertically at all before it hits the speed bump. So this gives the initial conditions at t = −π/(2V).) Call the solution of (2) y_{a}(t). Then, at t = π/(2V), we solve a second IVP:

(3)  \{ \begin{matrix}y′′ + y = 0 \\ y(π/(2V)) = y_{a}(π/(2V)) \\ y′(π/(2V)) = y′_{a}(π/(2V)) \end{matrix}

to get the continuation of the solution to t > π/(2V).

The first step is to find the function y_{a}(t) solving (2). The general solution of the homogeneous equation y′′ + y′ = 0 is y = c_{1} cos(t) + c_{2} sin(t). By the form of the right hand side of the ODE in (2), for V ≠ 1, we expect a particular solution of the form

y_{p} = A cos(V t) + B sin(V t).

Substituting into (2), we find that

A(1 − V^{2}) cos(V t) + B(1 − V^{2}) sin(V t) = cos(V t)

so A = \frac{1}{1−V^{2}} and B = 0.

Now, we determine the constants c_{1}, c_{2} in the general solution

y = c_{1} cos(t) + c_{2} sin(t) + \frac{1}{1 − V^{2}} cos(V t)

to get y(−π/(2V)) = 0 and y′(−π/(2V)) = 0. These initial conditions say:

c_{1} cos(−π/(2V)) + c_{2} sin(−π/(2V)) + 0 = 0
−c_{1} sin(−π/(2V)) + c_{2} cos(−π/(2V)) = \frac{V}{V^{2} − 1}

By Cramer’s Rule again,

c_{1} = det ( \begin{matrix}0 & sin(−π/(2V)) \\ \frac{V}{V^{2}−1} & cos(−π/(2V)) \end{matrix} ) ; det ( \begin{matrix}cos(−π/(2V)) & sin(−π/(2V)) \\ − sin(−π/(2V)) & cos(−π/(2V)) \end{matrix} )

= \frac{V sin(π/(2V))}{V^{2} − 1}

c_{2} = det ( \begin{matrix}cos(−π/(2V)) & 0 \\ − sin(−π/(2V)) & \frac{V}{V^{2}−1} \end{matrix} ) ; det ( \begin{matrix}cos(−π/(2V)) & sin(−π/(2V)) \\ − sin(−π/(2V)) & cos(−π/(2V)) \end{matrix} )

= \frac{V cos(π/(2V))}{V^{2} − 1}

This gives the coefficients in y_{a}(t), the formula for the solution on −π/(2V) < t < π/(2V).

Now, we want to solve the second IVP (3). If we call the solution y_{b}(t), then from the ODE in (3) – the homogeneous equation y′′ +y = 0, we know that y_{b} = d_{1} cos(t)+d_{2} sin(t), and we must determine d_{1}, d_{2} so that

y_{b}(π/(2V)) = y_{a}(π/(2V)) = c_{1} cos(π/(2V))+c_{2} sin(π/(2V)) = \frac{2V sin(π/(2V)) cos(π/(2V))}{V^{2} − 1}

and

y′_{b}(π/(2V)) = y′_{a}(π/(2V)) = \frac{2V cos^{2}(π/(2V))}{V^{2} − 1}.

So,

cos(π/(2V))d_{1} + sin(π/(2V))d_{2} = \frac{2V sin(π/(2V)) cos(π/(2V))}{V^{2} − 1}
− sin(π/(2V))d_{1} + cos(π/(2V))d_{2} = \frac{2V cos^{2}(π/(2V))}{V^{2} − 1}

Solving by Cramer’s Rule, we find d_{1} = 0, and d_{2} = \frac{2V cos(π/(2V))}{V^{2}−1}. Thus the solution on t ≥ π/(2V) is

y_{b}(t) = \frac{2V cos(π/(2V))}{V^{2} − 1} sin(t)

Plotting the amplitude as a function of V, we find A(V) = \frac{2V cos(π/(2V))}{V^{2}−1} has a maximum at roughly V ≐ .73.