Partial Solutions to 18.03 Problem Set 3

Part I 13.

33. y(x) = 0. 35. y(x) = c_{1}+c_{2}e^{−5x}. 37. y(x) = c_{1}e^{−x/2}. 39. y(x) = (c_{1}+c_{2}x)e^{−x/2}. (These are all bounded: use l’Hopital’s rule to check that lim_{x→∞} y(x) = 0.) 41. y(x) = c_{1}e^{−4x/3}.

Part II 12. This problem was revised so the Wronskian agreed with the definition given in the book (w = y_{1}y′_{2} − y′_{1}y_{2}). This is a solution to the original problem; for the revised problem, some signs must be changed. Full credit will be given to solutions of either version.

(a) w′ = (y′_{1}y_{2} − y′_{2}y_{1})′ = (y′_{1}y′_{2} + y_{2}y′′_{1}) − (y_{1}y′′_{2} − y′_{1}y′_{2}) = y′′_{1}y_{2} − y_{1}y′′_{2}.

(b)

w′ = y′′_{1}y_{2} − y_{1}y′′_{2}
= (−p(x)y′_{1} − q(x)y_{1})y_{2} − (−p(x)y′_{2} − q(x)y_{2})y_{1}
= −p(x)(y′_{1}y_{2} − y_{1}y′_{2})
= −p(x)w.

Hence w′ + p(x)w = 0.

(c) Exponentials are never zero. So if w(x) = ce^{− ∫ p(x)dx} is 0 for some x, then c is 0, so w(x) is zero for all x.

If p(x) is the 0 function, then w′ = −pw = 0, so w is constant.

(d) If a = 0, then two linearly independent solutions are y_{1} = 1 and y_{2} = x. The Wronskian is -1.

If a is positive, then two linearly independent solutions are y_{1} = sin(\sqrt{a}x) and y_{2} = cos(\sqrt{a}x). The Wronskian is

y′_{1}y_{2} − y_{1}y′_{2} = \sqrt{a} cos^{2}(\sqrt{a}x) + \sqrt{a} sin^{2}(\sqrt{a}x) = \sqrt{a}.

If a is negative, then two linearly independent solutions are y_{1} = e^{\sqrt{a}x} and y_{2} = e^{−\sqrt{a}x}. The Wronskian is

y′_{1}y_{2} − y_{1}y′_{2} = (\sqrt{a}e^{\sqrt{a}x}) e^{−\sqrt{a}x} − e^{\sqrt{a}x} (−\sqrt{a}e^{−\sqrt{a}x}) = 2\sqrt{a}.

(e) If y_{1} = x, then x^{2}y′′_{1} + xy′_{1} − y_{1} = 0 + x · 1 − x = 0 as desired.

In standard form, the differential equation is y′′ + \frac{1}{x}y′ − \frac{1}{x^{2}} y = 0, so p(x) = \frac{1}{x}. From Abel’s formula, the Wronskian w (using c = 1) is:

w = e^{− ∫ p(x)dx} = e^{− ln x} = \frac{1}{x}.

But also w = y′_{1}y_{2} − y_{1}y′_{2}, so \frac{1}{x} = y_{2} − xy′_{2}. We now know how to find y_{2}: rewrite this as y′_{2} − \frac{1}{x}y_{2} = \frac{1}{x^{2}} . Multiplying by an integrating factor of 1/x, we find (\frac{d}{dx} \frac{y_{2}}{x}) = \frac{1}{x}y′_{2} − \frac{1}{x^{2}} y_{2} = \frac{1}{x^{3}} . Hence \frac{y_{2}}{x} = c_{2} − \frac{1}{2x^{2}} , so y_{2} = c_{2}x − \frac{1}{2x}. For any c, this is linearly independent of y_{1} = x. With c = 0, we have −2/x. By linearity (and homogeneity), y_{2} = \frac{1}{x} is also a solution. Check: y′_{2} = −\frac{1}{x^{2}} , y′′_{2} = \frac{2}{x^{3}} , so

x^{2}y′′_{2} + xy′_{2} − y_{2} = x^{2} · \frac{2}{x^{3}} + x · (−\frac{1}{x^{2}} ) − \frac{1}{x} = 0.