Consider the following second order homogeneous differential equation!


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


Using product rule of differentiation we have


Know that

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




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





c is an arbitrary constant


$f_1$ & $f_2$ 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!


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

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


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

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 $W'(x)$:

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


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

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):

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.