Mathematics 304 – Ordinary Differential Equations Selected Solutions – Problem Set 3 October 25, 2000

§4.5/ 54.

a) The auxiliary equation of y′′ − s^{2}y = 0 is 0 = r^{2} − s^{2} = (r − s)(r + s), so by results proved in class, a fundamental set of solutions is {y_{1}, y_{2}} = {e^{sx}, e^{−sx}}, and the general solution looks like

y = c_{1}e^{sx} + c_{2}e^{−sx}

where c_{1}, c_{2} are arbitrary constants.

b) Each function c_{1}e^{sx} + c_{2}e^{−sx} can also be written as

c_{1}e^{sx} + c_{2}e^{−sx} = c_{1}(cosh(sx) + sinh(sx)) + c_{2}(cosh(sx) − sinh(sx))
= (c_{1} + c_{2}) cosh(sx) + (c_{1} − c_{2}) sinh(sx)

Conversely, every linear combination of cosh(sx), sinh(sx) can also be rewritten as:

d_{1} cosh(sx) + d_{2} sinh(sx) = d_{1} \frac{e^{sx} + e^{−sx}}{2} + d_{2} \frac{e^{sx} − e^{−sx}}{2}

= (\frac{d_{1} + d_{2}}{2} ) e^{sx} + (\frac{d_{1} − d_{2}}{2} ) e^{−sx}

Therefore d_{1} cosh(sx) + d_{2} sinh(sx) is also a general solution of y′′ − s^{2}y = 0, since this form produces exactly the same collection of functions as the form c_{1}e^{sx} + c_{2}e^{−sx} when c_{i} and d_{i} take on all possible real values.

c) To solve inital value problem for y′′ − s^{2}y = 0 with y(0) = a, y′(0) = b, we can use either form. The second form of the general solution, d_{1} cosh(sx) + d_{2} sinh(sx), is somewhat simpler. Note that cosh(0) = 1 and sinh(0) = 0 by definition. Hence from a = y(0) = d_{1} cosh(0) + d_{2} sinh(0) = d_{1}, we get d_{1} = a immediately. Then y′(0) = sd_{1} sinh(0) + sd_{2} cosh(0) = sd_{2} so the other initial condition gives b = y′(0) = sd_{2}, so d_{2} = b/s:

y(x) = a cosh(sx) + \frac{b}{s} sinh(sx)

§4.2/37. a) We assume that L(y) = y′′ + py′ + qy = 0 has p, q continuous on (a,b), and that ϕ is a solution of this ODE that is not identically zero, but has a non-isolated zero x_{0}. In other words, every deleted neighborhood of x_{0} contains another zero of ϕ: For all δ > 0, there exists some x ≠ x_{0} with |x − x_{0}| < δ such that ϕ(x) = 0. In particular for δ = 1/n, we get an x_{n} with |x_{n} − x_{0}| < 1/n such that ϕ(x_{n}) = 0. By the construction, the x_{n} form a sequence of real numbers converging to x_{0}. Now, in the definition of the derivative ϕ′(x_{0}):

ϕ′(x_{0}) = lim_{x→x_{0}} \frac{ϕ(x) − ϕ(x_{0})}{x − x_{0}},

we can let x → x_{0} through the terms in the x_{n} sequence. Since we require that ϕ′(x_{0}) exists, we should get the same limit no matter how x → x_{0}. But ϕ(x_{n}) = 0 for all n, so

ϕ′(x_{0}) = lim_{n→∞} \frac{ϕ(x_{n}) − ϕ(x_{0})}{x_{n} − x_{0}} = lim_{n→∞} \frac{0}{x_{n} − x_{0}} = 0.

Therefore L(ϕ) = 0, ϕ(x_{0}) = 0, and ϕ′(x_{0}) = 0.

b) The zero function also satisfies the IVP L(y) = 0, y(x_{0}) = 0, and y′(x_{0}) = 0. Therefore, since we assume p, q are continuous on an interval containing x_{0}, the Existence and Uniqueness Theorem (Theorem 2, page 165 of text) applies. Hence ϕ from part a) must be the zero function. But this is a contradiction, since we assumed that ϕ was not identically zero. Therefore, no solution of L(y) = 0 can have a non-isolated zero.

§4.3/4. We are given y_{1} = x^{2} cos(ln(x)), y_{2} = x^{2} sin(ln(x)). To decide whether these are linearly independent or not, we can consider a linear combination c_{1}y_{1}(x) + c_{2}y_{2}(x) ≡ 0 (i.e. the linear combination adds up to the zero function). Then we get another true equation by substituting any x ∈ (0, 1) in to this equation. For instance with x = e^{−π}, we get

0 = c_{1}y_{1}(e^{−π}) + c_{2}y_{2}(e^{−π})
= c_{1}e^{−2π} cos(−π) + c_{2}e^{−2π} sin(−π)
= −c_{1}e^{−2π}

(since sin(−π) = 0). Hence c_{1} = 0. Then substituting x = e^{−π/2}, we get c_{2} = 0 as well. {y_{1}, y_{2}} is linearly independent. The Wronskian works out to

W [y_{1}, y_{2}](x) = x^{3}

(using cos^{2}(ln(x)) + sin^{2}(ln(x)) = 1).

23. If y = ue^{− \frac{1}{2} ∫ p(x) dx}, then by the Chain and Product Rules for derivatives

\begin{aligned}\frac{dy}{dx} & = ue^{− \frac{1}{2} ∫ p(x) dx} (\frac{−1}{2}p(x)) + \frac{du}{dx}e^{− \frac{1}{2} ∫ p(x) dx} \\ & = e^{− \frac{1}{2} ∫ p(x) dx} (\frac{du}{dx} + \frac{−1}{2}p(x)) \end{aligned}

Continuing to compute the second derivative, again by Chain and Product Rules:

\frac{d^{2}y}{dx^{2}} = e^{− \frac{1}{2} ∫ p(x) dx} (\frac{d^{2}u}{dx^{2}} − p\frac{du}{dx} − \frac{1}{2} \frac{dp}{dx}u + \frac{1}{4}p^{2}u)

We substitute these into

0 = L(y) = \frac{d^{2}y}{dx^{2}} + p(x)\frac{dy}{dx} + q(x)y

and factor out the e^{− \frac{1}{2} ∫ p(x) dx} from all terms. The −p\frac{du}{dx} from \frac{d^{2}y}{dx^{2}} and the p\frac{du}{dx} from p\frac{dy}{dx} cancel leaving

0 = e^{− \frac{1}{2} ∫ p(x) dx} (\frac{d^{2}u}{dx^{2}} + (\frac{−1}{2}p′ − \frac{1}{4}p^{2} + q) u)

Multiplying by e^{\frac{1}{2} ∫ p(x) dx}, we get a second order ODE of the desired form for the function u:

0 = \frac{d^{2}u}{dx^{2}} + (\frac{−1}{2}p′ − \frac{1}{4}p^{2} + q) u

where f(x) = \frac{−1}{2}p′(x) − \frac{1}{4}p(x)^{2} + q(x).

31. We have L(y) = x^{2}y′′ − 3xy′ + 3y = 0 (a Cauchy-Euler equation).

a) y_{1} = x and y_{2} = x^{3} are solutions on (−∞, ∞) since

\begin{aligned}L(x) & = x^{2} · 0 − 3x · 1 + 3x = 0 \\ L(x^{3}) & = x^{2} · 6x − 3x · 3x^{2} + 3x^{3} = 0 \end{aligned}

The Wronskian of y_{1} and y_{2} is

W [y_{1}, y_{2}](x) = det ( \begin{matrix}x & x^{3} \\ 1 & 3x^{2} \end{matrix} ) = 2x^{3}

This is not zero (e.g.) at x = 1.

b) {y_{1}, y_{2}} is linearly independent, since if c_{1}y_{1} + c_{2}y_{2} = c_{1}x + c_{2}x^{3} ≡ 0, then we would get true equations by substituting any value for x. E.g., with x = 1 and x = 2, we get the equations

\begin{aligned}0 & = c_{1} + c_{2} \\ 0 & = 2c_{1} + 8c_{2} \end{aligned}

The only solution is c_{1} = c_{2} = 0, so these are linearly independent.

c) y_{3} = |x|^{3} is also a solution of L(y) = 0 on (−∞, ∞) since

y_{3}(x) = \{ \begin{matrix}x^{3} & if x ≥ 0 \\ −x^{3} & if x < 0 \end{matrix} ;  y′_{3}(x) = \{ \begin{matrix}3x^{2} & if x ≥ 0 \\ −3x^{2} & if x < 0 \end{matrix} ;  y′′_{3}(x) = \{ \begin{matrix}6x & if x ≥ 0 \\ −6x & if x < 0 \end{matrix}

Therefore

L(y_{3}) = \{ \begin{matrix}x^{2} · 6x − 3x · 3x^{2} + 3x^{3} = 0 & if x ≥ 0 \\ x^{2} · (−6x) − 3x · (−3x^{2}) − 3x^{3} = 0 & if x < 0 \end{matrix}

so y_{3} is also a solution defined on (−∞, ∞).

d) Is y_{3} = c_{1}y_{1} + c_{2}y_{2} for some c_{1}, c_{2}? The answer is no: Let x = 1; we get 1 = c_{1} + c_{2}. But with x = −1 we get 1 = −c_{1} − c_{2}. These equations are inconsistent. So there are no c_{1}, c_{2} ∈ R such that y_{3} = c_{1}y_{1} + c_{2}y_{2} for all x ∈ R. (Note that there are such constants, though, if we only take x > 0 or only x < 0.)

e) We are not seeing a contradiction to the uniqueness of solutions here since the theorem on existence and uniqueness of solutions does not apply on the interval (−∞, ∞). To write our equation in the standard form y′′ + py′ + qy = 0, we would have to divide through by x^{2}:

y′′ + (\frac{−3}{x}) y′ + (\frac{3}{x^{2}} ) y = 0

But then the p = \frac{−3}{x} and q = \frac{3}{x^{2}} functions have discontinuities at x = 0. The fact that there is not a unique solution with y(1) = 1, y′(1) = 3 is not a contradiction!