# 18.034 Practice Midterm 1 Answers

1. The general solution is y = sin t + C e^(-t^2), so the specific solution is y=sin t + e^(-t^2). The long-term behavior of the solution: a sine wave.

Warning! The earlier answer to 2 was backwards! Thanks to Justin for pointing this out.

2. Draw the phase line, which should look vaguely like this: >>><<<>>>. The equilibria are -1/3 (stable) and 1/3 (unstable). For large t (with y(0)=0), the particle moves to the left and tends to -1/3.

3. This is a separable equation. Solving (9 y^2-1)y'= 1, we get 3 y^3 - y = t+C. Substituting y(0)=0, we get C=0. Hence the solution satisfies 3 y^3-y = t. The graph is a cubic with one maximum and one minimum, turned on its side. If you sketch it, you will see why the solution doesn't exist for all time. (How long does the solution exist for?)

4. Solving the characteristic equation, we get x_n = A (1/2)^n + B (1/3)^n. Using the initial conditions, we find A=1, B=-1. For any initial conditions, it's clear that x_n --> 0.

5. 2 cos(3t + pi/3).

6. |z_1| = sqrt(a^2+b^2), |z_2| = sqrt(c^2+d^2), z_1 z_2 = (ac-bd) + (ad+bc)i, so |z_1 z_2| = sqrt( (ac-bd)^2 + (ad+bc)^2). You can expand and check that

(a^2 + b^2) (c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2.

7. f and (partial f/ partial y). The differential equation y' = y^(1/3) doesn't satisfy the hypotheses of the theorem, and also doesn't satisfy the conclusions: y=0 and y=2/3 t^(3/2) both satisfy the initial conditions y(0)=0.

8. The general solution is of the form R e^(-3t) cos(t - delta) or, if you prefer, (A cos t + B sin t) e^(-3t). A typical solution has sinusoidal behavior with exponential decay; it is underdamped. Changing the 6 (the damping) 2 root(10) will make it critically damped; making it larger will make it overdamped.