18.03 Hour Exam II: March 17, 1999: Solutions

[10] 1. (a) Find a particular solution (call it y_{1}) to y′′ + 2y′ + y = x^{2}.

Undetermined coefficients:

\begin{aligned}y & = & ax^{2} + bx + c \\ 2y′ & = & 4ax + 2b \\ y′′ & = & 2a \end{aligned}

Equating coefficients with x^{2} + 0x + 0, a = 1, b = −4, c = 6. So y_{1} = x^{2} − 4x + 6.

[10] (b) Find a particular solution (call it y_{2}) to y′′ + 2y′ + y = e^{3x} sin x.

The characteristic polynomial is f(r) = (r + 1)^{2}, and f(3 + i) = (4 + i)^{2} = 15 + 8i ≠ 0, so z′′ + 2z + z = e^{(3+i)x} has solution

z_{p} = \frac{e^{(3+i)x}}{15 + 8i} = \frac{(15 − 8i)e^{3x}(cos x + i sin x)}{15^{2} + 8^{2}}

whose imaginary part is

y_{2} = e^{3x} \frac{15 sin x − 8 cos x}{15^{2} + 8^{2}}.

[10] (c) Find a particular solution to y′′ + 2y′ + y = 2x^{2} + 3e^{3x} sin x in terms of y_{1} and y_{2} from parts (a) and (b). (You may leave your answer in terms of the symbols y_{1}, y_{2}!) 2y_{1} + 3y_{2}.

[10] 2. Find a pair of solutions y_{1}, y_{2}, to y′′ + 2y′ + y = 0 such that y_{1}(0) = 1, y′_{1}(0) = 0, y_{2}(0) = 0, y′_{2}(0) = 1. The repeated root leads to general solution ae^{−x} + bxe^{−x}, with derivative (−a + b)e^{−x} − bxe^{−x}. Evaluating at 0 gives a = 1, −a + b = 0 for y_{1} and a = 0, −a + b = 1 for y_{2}, so

y_{1} = (1 + x)e^{−x}, y_{2} = xe^{−x}.

[20] 3. y′′ +py′ +y = 0 represents a free spring-mass-dashpot system. We can set the dashpot constant p ≥ 0 as we like. Make a table showing what values of p lead to each of the following behaviors, and for each make a little sketch of a typical solution: Undamped oscillation, underdamping, overdamping. The characteristic polynomial has p^{2} − 4 in the square root, so: p = 0 leads to pure oscillation 0 < p < 2 leads to underdamping or damped oscillation p > 2 leads to overdamping (no oscillation)

4. Now we’ll set p = 1 in the previous equation and drive this system with a periodic force: y′′ + y′ + y = e^{iω_{0}x}.

[10] (a) What is the periodic or “steady state” solution? (You may leave your answer in terms of a complex exponential.)

The characteristic polynomial is f(r) = r^{2} + r + 1, and f(iω_{0}) = (1 − ω_{0}^{2}) + iω_{0} is not zero, so the periodic solution is

e^{iω_{0}x}/((1 − ω_{0}^{2}) + iω_{0}).

[5] (b) In terms of ω_{0}, what is the gain, or amplitude multiplier? (The gain is the the ratio of the amplitude of the periodic solution to the amplitude of the driving term (which is 1 in this case).)

The gain is 1/|(1 − ω_{0}^{2}) + iω_{0}| = 1/\sqrt{(1 − ω_{0}^{2})^{2} + ω_{0}^{2}} = 1/\sqrt{ω_{0}^{4} − ω_{0}^{2} + 1}.

[5] (c) For what value of ω_{0} is this gain the largest? (i.e., for what value of ω_{0} do we observe “practical resonance”?)

Maximum occurs when the quadratic in the denominator is a minimum, namely at the line of symmetry of the parabola, ω_{0}^{2} =“−b/2”= 1/2. So

ω_{0} = 1/\sqrt{2}.

[10] 5. (a) Assume that y = e^{2x}u and y′′ + 2y′ + y = e^{2x}q(x) for some function q(x). Write down an ODE satisfied by u which is free of exponential terms.

ESL: The characteristic polynomial is f(r) = (r + 1)^{2} again, so

e^{2x}q(x) = f(D)(e^{2x}u) = e^{2x}f(D + 2I)u.

Since f(r + 2) = (r + 3)^{2} = r^{2} + 6r + 9, the ODE for u is

u′′ + 6u′ + 9u = q(x).

[10] (b) Find the general solution to the inhomogeneous non-constant coefficient second order linear ordinary differential equation xy′′ + y′ = 1. (Hint: it’s reducible.)

Write v = y′, so xv′ + v = 1. This is separable: dv/(1 − v) = dx/x, or − ln |1 − v| = ln |x| + c, so 1 − v = a/x for a constant a: v = 1 − a/x. Integrating,

y = x − a ln |x| + b.

Friday’s lecture: music, jump ropes, and pancake stacks (EP 2.10)