Partial Solutions to 18.03 Problem Set 4, Part II

14. {r_{1}, r_{2}} = {1, 2}. If we take r_{1} = 1, r_{2} = 2, we get (D − I)u = x^{3}, i.e. u′ − u = x^{3}. The integrating factor is e^{−x}, so we get \frac{d}{dx}(ue^{−x}) = x^{3}e^{−x}. Integrating, we get ue^{−x} = e^{−x}(−x^{3} − 3x^{2} − 6x − 6) + C_{1}, so

u = −x^{3} − 3x^{2} − 6x − 6 + C_{1}e^{x}.

Then y′ − 2y = (D − 2I)y = −x^{3} − 3x^{2} − 6x − 6 + C_{1}e^{x}. The integrating factor is e^{−2x}, so we get

\frac{d}{dx}e^{−2x}y = (−x^{3} − 3x^{2} − 6x − 6)e^{−2x} + C_{1}e^{−x}.

Integrating and multiplying by e^{2x}, we get the full set of solutions:

y = \frac{1}{2}x^{3} + \frac{9}{4}x^{2} + \frac{21}{4}x + \frac{45}{8} + C_{2}e^{2x} − C_{1}e^{x}.

If we instead take r_{1} = 2, r_{2} = 1, we would have u′ − 2u = (D − 2I)u = x^{3}. The integrating factor is e^{−2x}, so we get \frac{d}{dx}(e^{−2x}u) = e^{−2x}x^{3}, from which

u = −\frac{1}{2}x^{3} − \frac{3}{4}x^{2} − \frac{3}{4}x − \frac{3}{8} + C_{1}e^{2x}.

Then y′ − y = u, so we use an integrating factor of e^{−x} to get:

\frac{d}{dx}e^{−x}y = e^{−x} (−\frac{1}{2}x^{3} − \frac{3}{4}x^{2} − \frac{3}{4}x − \frac{3}{8}) + C_{1}e^{x}.

Integrating and multiplying by e^{x}, we get

y = \frac{1}{2}x^{3} + \frac{9}{4}x^{2} + \frac{21}{4}x + \frac{45}{8} + C_{1}e^{2x} + C_{2}e^{x}.

15. (a) The solutions for the homogeneous case are of the form A cos ω_{0}t + B sin ω_{0}t. A specific solution to the original equation α cos ωt for some constant α. Solving for α: α(ω_{0}^{2} − ω^{2}) cos ωt = a cos ωt, so α = a/(ω_{0}^{2} − ω^{2}). Thus the general solution is

\frac{a}{ω_{0}^{2} − ω^{2}} cos ωT + A cos ω_{0}t + B sin ω_{0}t.

Notice that the solution stays bounded.

(b), (c) The particular solution is

\frac{a}{ω_{0}^{2} − ω^{2}} (cos ωt − cos ω_{0}t) = \frac{2a}{ω_{0}^{2} − ω^{2}} (sin \frac{(ω_{0} − ω)t}{2}) (sin \frac{(ω_{0} + ω)t}{2}) .

(d) We look for a solution of the form x = αt sin ωt for some α. Then ẍ + ω^{2}x = 2αω cos ωx, so α = \frac{a}{2ω}. Hence the general solution is

x = \frac{a}{2ω} t sin ωt + A cos ωt + B sin ωt.

16. (a) Rewrite \frac{dy}{dt} = y^{2} − a^{2} as

\frac{1}{2a} (\frac{1}{y − a} − \frac{1}{y + a}) dy = dt.

(We’ve lost the solutions y = a and y = −a; we’ll put them back in later.) Integrate to get

\frac{1}{2a} ln |(y − a)/(y + a)| = t + C.

Exponentiate to get

\frac{y − a}{y + a} = ±e^{2a(t+C)},

where C is any real number. By changing constants, we can rewrite this as

\frac{y − a}{y + a} = C_{2}e^{2at}.

Note that we’ve returned the “lost” solution y = a (the case C_{2} = 0). Solving for y, we get:

y = a\frac{1 + C_{2}e^{2at}}{1 − C_{2}e^{2at}}

(as well as the “lost” solution y = −a). On the other hand, dsolve gives us

y = −a\frac{e^{2ta−2C_{1}a} − 1}{1 + e^{2ta−2C_{1}a}}.

These are the same (let C_{2} = −e^{−2C_{1}a}), except dsolve has lost some solutions (for example, y = a, y = −a, and the case where C_{2} is positive)!

(b) The correct solution to problem 4 is

y = \frac{e^{−x^{2}}}{(\sqrt{π}/2)erf(x) + C} − x.

(The scaling factor was missing!) On the other hand, dsolve gives

y = −\frac{2xC_{1}e^{x^{2}} − x\sqrt{π}erf(x)e^{x^{2}} − 2}{e^{x^{2}}(2C_{1} − \sqrt{π}erf(x))}
= \frac{−2}{e^{x^{2}}(2C_{1} − \sqrt{π}erf(x))} − x
= e^{−x^{2}} \frac{−2}{2C_{1} − \sqrt{π}erf(x)} − x
= e^{−x^{2}} \frac{1}{(\sqrt{π}/2)erf(x) + C} − x

(where C = −C_{1}) as desired.

(c) y(2) is approximately -1.8447.

(d) y_{1} = e^{−x}(cos(4x) + \frac{1}{4} sin(4x)), y_{2} = \frac{1}{4}e^{−x} sin(4x).