18.034 Problem Set #1 solution

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1. (a) Verify that y = xa solves the differential equation x2y�� = 2y if the constant a satisfies the
equation a2 − a − 2 = 0. Thus get the two solutions x2 and x−1. Note that the first is valid on the
whole interval −∞ < x < ∞ but the second on −∞ < x < 0 or 0 < x < ∞ only. This behavior is typical for a broad class of linear homogeneous equations known as equations of Euler type. For this class the substitution y = xa always lead to an algebraic equation for a. (b) The equation x2y���� = 2y�� admits a solution y = xa, where a is a nonzero constant. What are the possible values of a? 2. Suppose a function y = f(x) satisfies the differential equation dy = 4y sin 2x dx and the initial condition y(π) = e. The purpose of this exercise is to find y(π/6). (a) Separate variables and integrate to obtain ln y = c − 2 cos 2x, y > 0. By use of the initial
condition show that c = 3, and then get y(π/6).
(b) The initial condition y(π) = e means that x = π corresponds to y = e. Likewise, x = π/6
corresponds to y = a, where a = y(π/6). Integrating between corresponding limits gives
� a dy � π/6
= 4 sin 2x dx. e y π
Evaluate the definite integrals and solve the resulting equation for a.
(c) If xdy + 3ydx = 0 and y(−π) = e you can’t find y(π). Why not?
3. Birkhoff-Rota, pp. 5, #3.
4. (a) Show that y1(x) = 0 and y2(x) = x3/2 are solutions for x ≥ 0 of the differential equation
y� = (3/2)y1/3.
(b) Discuss that all nonnegative solutions of the differential equation in part (a) starting at (0, 0)
lie between two solutions y1 and y2 and that the solution
0 for x < c, y(x) = (x − c)3/2 for x ≥ c fills out the funnel between them. 5. Birkhoff-Rota, pp. 11, #7. (Typo. k is n.) 6. (The Bernoulli equation.) It is a differential equation of the form y� + p(x)y = q(x)yn with n =� 1. (a) Show that it becomes linear by the change of variable∗ u = y1−n. (Hint. Begin by dividing both sides of the equation by yn.) (b) Solve the Bernoulli equation y� + y = xy3 using the method in part (a). ∗This trick was found by Leibniz in 1696. 1 .