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Consider the following second order homogeneous differential equation!

$$a_0(x)\frac{d^2y}{dx^2}+a_1(x)\frac{dy}{dx}+a_2(x)y=0$$

Suppose that $a_0,a_1,a_2$ are continuous for all values of $[a,b]$. Let $f_1$ and $f_2$ be two distinct solutions to the above differential equation for all $x$ on $a \leq x \leq b$. Further suppose that $f_2(x)\neq 0$. Let $W[f_1(x),f_2(x)]$ be Wronskian of $f_1$ & $f_2$ at $x$.

  1. Show that $$\frac{\text{d}}{\text{d}x}\left[\frac{f_1(x)}{f_2(x)}\right]=-\frac{W[f_1(x),f_2(x)]}{[f_2(x)]^2}$$

My work is as follows

$$y=\frac{f_1(x)}{f_2(x)}$$

Using product rule of differentiation we have

$$\frac{\text{d}y}{\text{d}x}=\frac{f_2(x)f_1'(x)-f_2'(x)f_1(x)}{[f_2(x)]^2}$$

Know that

$$W[f_1(x),f_2(x)]=|\begin{bmatrix}f_1(x) & f_2(x) \\f_1'(x) & f_2'(x) \end{bmatrix}|$$

$$W[f_1(x),f_2(x)]=f_1(x)f_2'(x)-f_1'(x)f_2(x)$$

$$-W[f_1(x),f_2(x)]=-[f_1'(x)f_2(x)-f_1(x)f_2'(x)]$$

$$\frac{\text{d}y}{\text{d}x}=-\frac{W[f_1(x),f_2(x)]}{[f_2(x)]^2}$$

for $x$ on $[a,b]$

Is is that simple?

  1. Use result in part 1 to show that if $W[f_1(x),f_2(x)]=0$ then, $f_1$ & $f_2$ are linearly dependent!

My attempt

Assume that $W[f_1(x),f_2(x)]=0$

$$\frac{\text{d}}{\text{d}x}[\frac{f_1(x)}{f_2(x)}]=-\frac{W[f_1(x),f_2(x)]}{[f_2(x)]^2}$$

$$\int{\text{d}}\left[\frac{f_1(x)}{f_2(x)}\right]=-\int\frac{W[f_1(x),f_2(x)]}{[f_2(x)]^2}{\text{d}x}$$

$$\int{\text{d}}\left[\frac{f_1(x)}{f_2(x)}\right]=-\int0{d}x$$

$$\frac{f_1(x)}{f_2(x)}=c$$

c is an arbitrary constant

$${f_1(x)}=c{f_2(x)}$$

$f_1$ & $f_2$ just differ by a constant. Hence they are linearly dependent!

Is this correct?

  1. Suppose that solution $f_1$ & $f_2$ are linearly independent on x on $a \leq x \leq b$ Hence, $f(x)=\frac{f_1(x)}{f_2(x)}$ and show that f is a monotonic function on $a \leq x \leq b$

I try to argue that The derivative of $f(x)$ is negative therefore monotonic!

I am totally unsure for my work. I sincerely hope that someone will provide the rigorous and correct way of doing this question!

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I am going to assume $a_0(x) \ne 0$ for $x \in [a, b]$; otherwise the given differential equation is singular and $y''(x)$ if much more difficult to meaningfully determine. Having said this:

The answer for item (1) given in the text of the question appears to be fine in essence, with the small and easily repaired error occurring in the equation

$-W[f_1(x),f_2(x)] = -[f_1'(x)f_2(x)-f_1(x)f_2'(x)],\tag{1}$

which should read

$-W[f_1(x),f_2(x)] = [f_1'(x)f_2(x)-f_1(x)f_2'(x)]; \tag{2}$

if this correction is adopted, the stated formula for $dy/dx$ readily follows; it is indeed that simple.

As for part (2), I think the argument can be clarified somewhat by providing limits for the integration operations performed, e.g., we might want to write

$\dfrac{f_1(x)}{f_2(x)} - \dfrac{f_1(a)}{f_2(a)} = \displaystyle \int_a^x \dfrac{d}{ds} \dfrac{f_1(s)}{f_2(s)}ds = \displaystyle \int_a^x 0ds = 0; \tag{3}$

thus

$\dfrac{f_1(x)}{f_2(x)} = \dfrac{f_1(a)}{f_2(a)} \tag{4}$

for $x \in [a, b]$; this approach not only shows that $f_1(x)/f_2(x)$ is a constant $c$, but also supplies $c$ with the value

$c = \dfrac{f_1(a)}{f_2(a)}; \tag{5}$

of course, the linear dependence of $f_1(x)$, $f_2(x)$ follows from (5) and (6), since they imply

$f_1(x) = \dfrac{f_1(a)}{f_2(a)}f_2(x) = cf_2(x) \tag{6}$

for all $x \in [a, b]$.

Turning to item (3), we compute $W'(x)$:

$W'(x) = (f_1(x) f_2'(x) - f_1'(x)f_2(x))' = f_1'(x) f_2'(x) + f_1(x) f_2''(x) - f_1''(x)f_2(x) - f_1'(x)f_2'(x) = f_1(x)f_2''(x) - f_1''(x) f_2(x); \tag{7}$

thus

$a_0(x)W'(x) = a_0(x)f_2''(x)f_1(x) - a_0(x)f_1''(x)f_2(x); \tag{8}$

we may now use the hypothesis that $f_1(x)$, $f_2(x)$ satisfy

$a_0(x) y''(x) + a_1(x) y'(x) + a_2(x) y(x) = 0 \tag{9}$

to eliminate the second derivatives from (8):

$a_0(x)W'(x) = -(a_1(x) f_2'(x) + a_2(x) f_2(x))f_1(x) + (a_1(x)f_1'(x) + a_2(x)f_1(x))f_2(x) = a_1(x)(f_1'(x)f_2(x) - f_2'(x)f_1(x)) = -a_1(x)W(x), \tag{10}$

whence

$W'(x) = - \dfrac{a_1(x)}{a_0(x)} W(x). \tag{11}$

The equation (11) has a simple solution

$W(x) = \exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) W(a) \tag{12}$

which holds for $x \in [a, b]$. Since

$\exp(- \displaystyle \int_a^x\dfrac{a_1(s)}{a_0(s)}ds) > 0,\tag{13}$

the sign of $W(x)$ cannot change on $[a, b]$; and since

$f'(x) = -\dfrac{W(x)}{(f_2(x))^2}, \tag{14}$

$f'(x)$ is also of fixed sign on $[a, b]$; thus $f(x)$ must indeed be monotonic on this interval.

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