Section 5.4 continued Solve x^{2}y′′ − 2xy′ = 0 (*). We could solve by letting v = y′, but we will instead use 5.4 methods Note x is an ordinary point iff x ≠ 0 (y′′ − \frac{2}{x}y′ = 0.) x = 0 is a singular point. Note x^{2}x^{r−2}r(r − 1) − 2xx^{r−1}r = 0 implies r^{2} − r − 2r = 0 and recall y = (−x)^{r} gives same equation for r as y = x^{r}. Thus y = |x|^{r} implies r^{2} + (α − 1)r + β = r^{2} − 3r + 0 = r(r − 3) = 0 Thus r = 0, 3. Thus y = |x|^{0} = 1 and y = |x|^{3} are solutions to (*) Since (*) is a linear equation, the general solution is y = c_{1} + c_{2}|x|^{3}. Note an equivalent general solution is y = k_{1} + k_{2}x^{3}. Both forms are valid for all x. When is a unique solution to the following initial value problem guaranteed?

x^{2}y′′ − 2xy′ = 0, y(t_{0}) = y_{0}, y′(t_{0}) = y_{1} (**)

y′′ − \frac{2}{x}y′ = 0, y(t_{0}) = y_{0}, y′(t_{0}) = y_{1}

Since \frac{2}{x} and the zero constant function are continuous on

(−∞, 0) ∪ (0, ∞),

(**) has a unique solution for t_{0} < 0 and this solution exists on

(−∞, 0).

(**) has a unique solution for t_{0} > 0 and this solution exists on

(0, ∞).

There are an infinite number of solutions for y(0) = a, y′(0) = 0.

How is x^{r} defined: If n is a positive integer: x^{n} = x · x · ... · x If m is a positive integer: If f(x) = x^{m}, then f^{−1}(x) = x^{\frac{1}{m}} and x^{\frac{n}{m}} = (x^{n})^{\frac{1}{m}} Let r ≥ 0. Let r_{n} be any sequence consisting of positive rational numbers such that lim_{n→∞}r_{n} = r. Then

x^{r} = lim_{n→∞}x^{r_{n}}.

See more advanced class for why the above is well-defined. If r < 0, then x^{r} = x^{−r}. If x is a real number, when is x^{r} a real number? x^{n} = x · x · ... · x is a real number when n is a positive integer. If f(x) = x^{n}, then the image of f = \{ \begin{matrix}real numbers & n odd \\ [0, ∞) & n even \end{matrix}

Thus if f^{−1}(x) = x^{\frac{1}{n}} is real-valued, then the domain of f^{−1} is \{ \begin{matrix}real numbers & n odd \\ [0, ∞) & n even \end{matrix}

In complex analysis, (\frac{1+i\sqrt{3}}{2})^{3} = −1, (−1)^{3} = −1, (\frac{1−i\sqrt{3}}{2})^{3} = −1

Recall (e^{\frac{iπ}{3}} )^{3} = (cos\frac{π}{3} + isin\frac{π}{3} )^{3} = −1 Complex numbers are also roots of unity:

(e^{\frac{2iπ}{3}} )^{3} = 1 (e^{\frac{−2iπ}{3}} )^{3} = 1, (1)^{3} = 1 Solve x^{2}y′′ + αxy′ + βy = 0. Let y = x^{r}, y′ = rx^{r−1}, y′′ = r(r − 1)x^{r−2} (case when y = (−x)^{r} is similar). x^{2}x^{r−2}r(r − 1) + αxx^{r−1}r + βx^{r} = 0 x^{r}[r^{2} − r + αr + β] = 0 for all x implies r^{2} + (α − 1)r + β = 0 Thus x^{r} is a solution iff r = ^{\frac{−(α−1)±\sqrt{(α−1)^{2}−4β}}{2}} Case 1: Two real roots, r_{1}, r_{2}. General solution is y = c_{1}|x|^{r_{1}} + c_{2}|x|^{r_{2}} Case 2: Two complex roots, r_{i} = λ ± iμ : Convert solution to form without complex numbers. Note |x|^{±iμ} = e^{ln(|x|^{±iμ})} = e^{(±iμ)ln|x|} = e^{i(±μln|x|)}

= cos(±μln|x|) + isin(±μln|x|) = cos(μln|x|) ± isin(μln|x|)

General solution is y = c_{1}|x|^{r_{1}} + c_{2}|x|^{r_{2}} = c_{1}|x|^{λ+iμ} + c_{2}|x|^{λ−iμ} = |x|^{λ}(c_{1}|x|^{iμ} + c_{2}|x|^{−iμ}) = |x|^{λ}(c_{1}[cos(μln|x|)+isin(μln|x|)]+c_{2}[cos(μln|x|)−isin(μln|x|)]) = |x|^{λ}([c_{1} + c_{2}]cos(μln|x|) + i[c_{1} − c_{2}]sin(μln|x|)) = |x|^{λ}(k_{1}cos(μln|x|) + k_{2}sin(μln|x|)) = k_{1}|x|^{λ}cos(μln|x|) + k_{2}|x|^{λ}sin(μln|x|) Case 3: one repeated root, r_{1} = \frac{−(α−1)}{2}. (i.e., \sqrt{(α − 1)^{2} − 4β} = 0): Thus |x|^{r_{1}} is a solution. Find 2nd solution.

Method 1. Reduction of order: Suppose y = u(x)|x|^{r_{1}} is a solution to x^{2}y′′ + αxy′ + βy = 0. Plug in and determine u(x) Method 2: Let L(y) = x^{2}y′′ + αxy′ + βy where y′ = \frac{dy}{dx}. L(|x|^{r}) = |x|^{r}(r − r_{1})^{2} \frac{∂}{∂r} [L(|x|^{r})] = \frac{∂}{∂r} [|x|^{r}(r − r_{1})^{2}] = (|x|^{r})′(r − r_{1})^{2} + 2|x|^{r}(r − r_{1}) = 0

if r = r_{1}.

Suppose x is constant with respect to r and all the partial derivatives are continuous. Then \frac{∂}{∂r} [L(y)] = \frac{∂}{∂r} [x^{2}y′′ + αxy′ + βy] = x^{2} \frac{∂y′′}{∂r} + αx\frac{∂y′}{∂r} + β \frac{∂y}{∂r}

= x^{2} \frac{∂}{∂r} [\frac{∂^{2}y}{∂x^{2}} ] + αx\frac{∂}{∂r} [\frac{∂y}{∂x}] + β \frac{∂y}{∂r} = x^{2} \frac{∂^{2}}{∂x^{2}} [\frac{∂y}{∂r}] + αx\frac{∂}{∂x}[\frac{∂y}{∂r}] + β \frac{∂y}{∂r}

= L(\frac{∂y}{∂r}) for all r

L(\frac{∂|x|^{r}}{∂r}) = \frac{∂}{∂r} [L(|x|^{r})] = 0 for r = r_{1}. \frac{∂|x|^{r}}{∂r} = \frac{∂e^{ln|x|^{r}}}{∂r} \frac{∂e^{rln|x|}}{∂r} = (e^{rln|x|})ln|x| = |x|^{r}ln|x| Thus |x|^{r_{1}}ln|x| is a solution. Thus general solution is y = c_{1}|x|^{r_{1}} + c_{2}|x|^{r_{1}}ln|x| since by the Wronskian, |x|^{r_{1}} and |x|^{r_{1}}ln|x| are linearly independent. Suppose x > 0 and r_{1} ≠ 0.

| \begin{matrix}x^{r_{1}} & x^{r_{1}}ln|x| \\ r_{1}x^{r_{1}−1} & r_{1}x^{r_{1}−1}ln|x| + x^{r_{1}−1} \end{matrix} |
= x^{r_{1}}(r_{1}x^{r_{1}−1}ln|x| + x^{r_{1}−1}) − x^{r_{1}}ln|x|r_{1}x^{r_{1}−1}

= x^{2r_{1}−1}[r_{1}ln|x| + 1 − ln|x|r_{1}] = x^{2r_{1}−1} ≠ 0 for x ≠ 0

Other cases for Wronskian are similar.