# Proofs about the second-order linear differential equation?

• Math100
Math100
Homework Statement
Let ## u_{1} ## and ## u_{2} ## be two linearly independent solutions of the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
(i) Let ## w=u_{2}/u_{1} ##. Show that ## w'=c/u_{1}^2 ## for some nonzero constant ## c ##.
(ii) Let ## y=-2u_{1}'/u_{1} ##. Show that ## y ## satisfies the Ricatti equation ## y'-\frac{y^2}{2}=f ##.
(iii) Show that ## w''/w'=-2u_{1}'/u_{1} ##, and prove that ## w ## satisfies the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=f ##.
(iv) Hence find a general solution of the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=2 ##. [You are not required to prove that your answer is the general solution.]
Relevant Equations
None.
Proof:

(i) Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u''+\frac{f}{2}u=0\implies r^2+\frac{f}{2}=0 ##, so ## r=\pm \sqrt{\frac{f}{2}}i ##.
This implies ## u_{1}=c_{1}cos(\sqrt{\frac{f}{2}}x) ## and ## u_{2}=c_{2}sin(\sqrt{\frac{f}{2}}x) ##.
Note that ## u_{1}'=-\sqrt{\frac{f}{2}}c_{1}sin(\sqrt{\frac{f}{2}}x), u_{2}'=\sqrt{\frac{f}{2}}c_{2}cos(\sqrt{\frac{f}{2}}x) ##.
Let ## w=\frac{u_{2}}{u_{1}} ##.
Then ## w'=\frac{u_{1}u_{2}'-u_{2}u_{1}'}{u_{1}^2}=\frac{c}{u_{1}^2} ##.
Observe that ## w'=\frac{c_{1}cos(\sqrt{\frac{f}{2}}x)\cdot \sqrt{\frac{f}{2}}c_{2}cos(\sqrt{\frac{f}{2}}x)-c_{2}sin(\sqrt{\frac{f}{2}}x)[-\sqrt{\frac{f}{2}}c_{1}sin(\sqrt{\frac{f}{2}}x)]}{u_{1}^2}=\frac{c}{u_{1}^2} ##.
Thus ## w'=\frac{\sqrt{\frac{f}{2}}c_{1}c_{2}cos^2(\sqrt{\frac{f}{2}}x)+\sqrt{\frac{f}{2}}c_{1}c_{2}sin^2(\sqrt{\frac{f}{2}}x)}{u_{1}^2}=\frac{c}{u_{1}^2} ##,
so ## w'=\frac{\sqrt{\frac{f}{2}}c_{1}c_{2}}{u_{1}^2}=\frac{c}{u_{1}^2} ##, where ## c=\sqrt{\frac{f}{2}}c_{1}c_{2} ##.
Therefore, ## w'=c/u_{1}^2 ## for some nonzero constant ## c ##.

(ii) Let ## y=-\frac{2u_{1}'}{u_{1}} ##.
Then ## y'=\frac{u_{1}(-2u_{1}'')-(-2u_{1}')(u_{1}')}{u_{1}^2}=\frac{-2u_{1}u_{1}''+2(u_{1}')^2}{u_{1}^2} ##.
Consider the Ricatti equation ## y'-\frac{y^2}{2}=f ##.
Observe that ## \frac{-2u_{1}u_{1}''+2(u_{1}')^2}{u_{1}^2}-\frac{1}{2}(-\frac{2u_{1}'}{u_{1}})^2=f ##
## \frac{-2u_{1}u_{1}''+2(u_{1}')^2}{u_{1}^2}-\frac{1}{2}(\frac{4(u_{1}')^2}{u_{1}^2}=f ##
## \frac{-2u_{1}u_{1}''+2(u_{1}')^2-2(u_{1}')^2}{u_{1}^2}=f ##
## \frac{-2u_{1}u_{1}''}{u_{1}^2}=f ##.
Since ## y'-\frac{y^2}{2}=\frac{-2u_{1}u_{1}''}{u_{1}^2}=f ##, it follows that ## -2u_{1}u_{1}''=fu_{1}^2\implies -2u_{1}''=fu_{1}\implies u_{1}''=-\frac{fu_{1}}{2}\implies u_{1}''+\frac{fu_{1}}{2}=0 ##.
Therefore, ## y ## satisfies the Ricatti equation ## y'-\frac{y^2}{2}=f ##.

(iii) Let ## w'=\frac{c}{u_{1}^2} ## for some nonzero constant ## c ##.
Then ## w''=(cu_{1}^{-2})'=-2cu_{1}^{-3}u_{1}'=-\frac{2cu_{1}'}{u_{1}^3} ## by the chain rule.
Thus ## \frac{w''}{w'}=\frac{-2cu_{1}'}{u_{1}^3}\cdot \frac{u_{1}^2}{c}=\frac{-2u_{1}'}{u_{1}} ##.
Therefore, ## y=\frac{-2u_{1}'}{u_{1}}=\frac{w''}{w'} ## (by part ii) and ## w ## satisfies the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=f ##.

(iv) By parts (ii) and (iii), we have that ## y=\frac{w''}{w'}=\frac{-2u_{1}'}{u_{1}} ##.
Then ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=2\implies y'-\frac{1}{2}y^2=2 ##.
Now we have ## \frac{dy}{dx}=2+\frac{y^2}{2} ##.
Observe that ## \frac{dy}{dx}=\frac{4+y^2}{2} ##
## \frac{dy}{4+y^2}=\frac{dx}{2} ##
## \int \frac{dy}{4+y^2}=\int \frac{dx}{2} ##
## \frac{1}{2}\cdot tan^{-1}(\frac{y}{2})=\frac{x}{2}+c ##
## tan^{-1}(\frac{y}{2})=x+c ##
## \frac{y}{2}=tan(x+c) ##
## y=2tan(x+c) ##.
Therefore, a general solution of the equation ## (\frac{w''}{w'})'-\frac{1}{2}(\frac{w''}{w'})^2=2 ## is ## y=2tan(x+c) ##.

I am assuming you are trying to find out whether your work is correct or not. If ##f(x)## is non-constant then your solution doesn't work. See why by substituting any non-constant function, x for example.

Mark44
mathhabibi said:
I am assuming you are trying to find out whether your work is correct or not. If ##f(x)## is non-constant then your solution doesn't work. See why by substituting any non-constant function, x for example.
Are you talking about the general solution? To which part of the problem are you referring to?

Math100 said:
Are you talking about the general solution? To which part of the problem are you referring to?
At the very beginning, you say that ##u_1=c_1\cos(\sqrt{\frac{f}2}x)##. I was saying that this is true only when ##f## is constant.

mathhabibi said:
At the very beginning, you say that ##u_1=c_1\cos(\sqrt{\frac{f}2}x)##. I was saying that this is true only when ##f## is constant.
Okay, I see that now. Then what's the error here? Is the characteristic equation wrong?

Math100 said:
Okay, I see that now. Then what's the error here? Is the characteristic equation wrong?
Your solutions ##u_1## and ##u_2## are generally incorrect. If ##f## is constant then you're fine.

Math100
I would write ##u_i =-\dfrac{2}{f}u_i''## so ##\omega =\dfrac{u_2}{u_1}=\dfrac{u_2''}{u_1''}## and start with both differentiations of ##\omega ## since you have to do this anyway.

My comments apply to part (i) only.
Math100 said:
Homework Statement: Let ## u_{1} ## and ## u_{2} ## be two linearly independent solutions of the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
(i) Let ## w=u_{2}/u_{1} ##. Show that ## w'=c/u_{1}^2 ## for some nonzero constant ## c ##.
<snip>

(i) Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u''+\frac{f}{2}u=0\implies r^2+\frac{f}{2}=0 ##, so ## r=\pm \sqrt{\frac{f}{2}}i ##.
1) Since f is a function, you should write this explicitly as f(x). Not doing so possibly led to the error that @mathhabibi points out.
2) You apparently are assuming a solution of the form ##y = e^{rx}##. This is something that you should state.
Math100 said:
This implies ## u_{1}=c_{1}cos(\sqrt{\frac{f}{2}}x) ## and ## u_{2}=c_{2}sin(\sqrt{\frac{f}{2}}x) ##.
Note that ## u_{1}'=-\sqrt{\frac{f}{2}}c_{1}sin(\sqrt{\frac{f}{2}}x), u_{2}'=\sqrt{\frac{f}
The error pointed out by @mathhabibi is that both ##u_1(x)## and ##u_2(x)## are functions that involve f(x), when you differentiate them, you need to use the chain rule.

Math100 and mathhabibi
Mark44 said:
My comments apply to part (i) only.
1) Since f is a function, you should write this explicitly as f(x). Not doing so possibly led to the error that @mathhabibi points out.
2) You apparently are assuming a solution of the form ##y = e^{rx}##. This is something that you should state.
The error pointed out by @mathhabibi is that both ##u_1(x)## and ##u_2(x)## are functions that involve f(x), when you differentiate them, you need to use the chain rule.
I noticed the issue. Since both of the functions ## u_{1}(x), u_{2}(x) ## were incorrect, then how should I solve this differential equation and find these correct functions?

mathhabibi said:
Your solutions ##u_1## and ##u_2## are generally incorrect. If ##f## is constant then you're fine.
I forgot the fact that ## f ## is a function, not a constant.

Math100 said:
I forgot the fact that ## f ## is a function, not a constant.
A week or two ago, I was trying to solve the first differential equation in your post, except that ##p=\frac f2##. According to Wolfram alpha, it is a Riccati differential equation, so it is probably unsolvable. If you find a solution to it, then get ready for fame.

Math100
fresh_42 said:
I would write ##u_i =-\dfrac{2}{f}u_i''## so ##\omega =\dfrac{u_2}{u_1}=\dfrac{u_2''}{u_1''}## and start with both differentiations of ##\omega ## since you have to do this anyway.
I see where you got the ## -\frac{2}{f}u_{i}''=u_{i} ## from. But I don't understand why/how does ## w=\frac{u_{2}}{u_{1}}=\frac{u_{2}''}{u_{1}''} ##. And by starting both differentiations of ## \omega ##, do you mean to take up to second derivatives of ## \omega ##?

Math100 said:
I see where you got the ## -\frac{2}{f}u_{i}''=u_{i} ## from. But I don't understand why/how does ## w=\frac{u_{2}}{u_{1}}=\frac{u_{2}''}{u_{1}''} ##. And by starting both differentiations of ## \omega ##, do you mean to take up to second derivatives of ## \omega ##?
##\omega =u_2/u_1## is a definition, and ## -\frac{2}{f}u_{i}''=u_{i} ## is the fact that they solve the differential equation. I only substituted the two into the definition of ##\omega ##. Of course, it could be that ##f(x)=0## somewhere, but I think we should first tackle the problem generically and bother about exceptions later on. Differentiating ##\omega ## yields new equations in which we could substitute former ones.

I don't think that you should try to solve the equation, especially as we don't know anything about ##f(x).## And it is not explicitly required. You are only asked about certain behaviors of flows (solutions). You can solve differential equations if you encounter easy ones e.g. ## v''+v=0## or ##v=v'.##

The general route to the proof should be:
- use as many pieces of information as you have
- calculate everything you have to calculate anyway at the beginning (for later use)
- ignore exceptions like ##f(x)=0## and deal with them at the end
- ##\omega ## is the quotient of two solutions; use this to get rid of common factors like ##-2/f.##
- draw solutions if necessary

Math100
I'm a little stale in this area, but, have you considered using characteristic curves?

WWGD said:
I'm a little stale in this area, but, have you considered using characteristic curves?
I'm working on this proof, almost done.

WWGD and fresh_42
WWGD said:
I'm a little stale in this area, but, have you considered using characteristic curves?
Yes, writing an article about the meaning of differential equations in general was a lot easier.

fresh_42 said:
##\omega =u_2/u_1## is a definition, and ## -\frac{2}{f}u_{i}''=u_{i} ## is the fact that they solve the differential equation. I only substituted the two into the definition of ##\omega ##. Of course, it could be that ##f(x)=0## somewhere, but I think we should first tackle the problem generically and bother about exceptions later on. Differentiating ##\omega ## yields new equations in which we could substitute former ones.

I don't think that you should try to solve the equation, especially as we don't know anything about ##f(x).## And it is not explicitly required. You are only asked about certain behaviors of flows (solutions). You can solve differential equations if you encounter easy ones e.g. ## v''+v=0## or ##v=v'.##

The general route to the proof should be:
- use as many pieces of information as you have
- calculate everything you have to calculate anyway at the beginning (for later use)
- ignore exceptions like ##f(x)=0## and deal with them at the end
- ##\omega ## is the quotient of two solutions; use this to get rid of common factors like ##-2/f.##
- draw solutions if necessary
Okay, here's my revised proof on part (i).

Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u_{i}''+\frac{fu}{2}=0\implies u_{i}''=-\frac{f}{2}u ##, so ## -\frac{2}{f}u_{i}''=u_{i} ##.
Given that ## w=\frac{u_{2}}{u_{1}} ##, we have ## w'=\frac{u_{1}u_{2}'-u_{2}u_{1}'}{u_{1}^2} ##.
Since ## u_{1} ## and ## u_{2} ## are linearly independent, it follows that their Wronskian is nonzero.
Note that ## W=u_{1}u_{2}'-u_{2}u_{1}'\neq 0 ##.
This implies ## W'=(u_{1}u_{2}'-u_{2}u_{1}')'=u_{1}u_{2}''+u_{1}'u_{2}'-u_{2}u_{1}''-u_{2}'u_{1}'=u_{1}u_{2}''-u_{2}u_{1}'' ##.
Observe that ## W'=u_{1}u_{2}''-u_{2}u_{1}''=u_{1}(-\frac{1}{2}f\cdot u_{2})-u_{2}(-\frac{1}{2}f\cdot u_{1})=0 ##.
Thus, ## W=\frac{u_{2}}{u_{1}}=u_{1}u_{2}'-u_{2}u_{1}' ## is a nonzero constant ## c ##.
Therefore, ## w'=\frac{c}{u_{1}^2} ## for some nonzero constant ## c ##.

fresh_42
WWGD said:
I'm a little stale in this area, but, have you considered using characteristic curves?
What are characteristic curves?

Math100 said:
Okay, here's my revised proof on part (i).

Consider the second-order linear differential equation ## \frac{d^2u}{dx^2}+\frac{fu}{2}=0, f=f(x) ##.
Then ## u_{i}''+\frac{fu}{2}=0\implies u_{i}''=-\frac{f}{2}u ##, so ## -\frac{2}{f}u_{i}''=u_{i} ##.
Given that ## w=\frac{u_{2}}{u_{1}} ##, we have ## w'=\frac{u_{1}u_{2}'-u_{2}u_{1}'}{u_{1}^2} ##.
Since ## u_{1} ## and ## u_{2} ## are linearly independent, it follows that their Wronskian is nonzero.
Note that ## W=u_{1}u_{2}'-u_{2}u_{1}'\neq 0 ##.
This implies ## W'=(u_{1}u_{2}'-u_{2}u_{1}')'=u_{1}u_{2}''+u_{1}'u_{2}'-u_{2}u_{1}''-u_{2}'u_{1}'=u_{1}u_{2}''-u_{2}u_{1}'' ##.
Observe that ## W'=u_{1}u_{2}''-u_{2}u_{1}''=u_{1}(-\frac{1}{2}f\cdot u_{2})-u_{2}(-\frac{1}{2}f\cdot u_{1})=0 ##.
Thus, ## W=\frac{u_{2}}{u_{1}}=u_{1}u_{2}'-u_{2}u_{1}' ## is a nonzero constant ## c ##.
Therefore, ## w'=\frac{c}{u_{1}^2} ## for some nonzero constant ## c ##.
The line before the last should be
Thus, ## W=u_{1}u_{2}'-u_{2}u_{1}' ## is a nonzero constant ## c ##.
since it is only the numerator of ##\omega .##

Math100
fresh_42 said:
The line before the last should be

since it is only the numerator of ##\omega .##
And also ## w ## instead of the Wronskian ## W ##.

Math100 said:
And also ## w ## instead of the Wronskian ## W ##.
And you didn't use a division by ##f(x)## so it should work in all cases where ##f(x)## has zeros, too!

Math100
fresh_42 said:
Yes, writing an article about the meaning of differential equations in general was a lot easier.
Any Weak/Distributional solutions?

## What is a second-order linear differential equation?

A second-order linear differential equation is a differential equation involving a function and its derivatives up to the second order. It can be written in the form $$a(x)y'' + b(x)y' + c(x)y = f(x)$$, where $$y$$ is the unknown function, $$y'$$ and $$y''$$ are its first and second derivatives, respectively, and $$a(x)$$, $$b(x)$$, $$c(x)$$, and $$f(x)$$ are given functions of $$x$$.

## What is the general solution to a homogeneous second-order linear differential equation?

The general solution to a homogeneous second-order linear differential equation $$a(x)y'' + b(x)y' + c(x)y = 0$$ is a linear combination of two linearly independent solutions. If $$y_1(x)$$ and $$y_2(x)$$ are these solutions, the general solution is $$y(x) = C_1 y_1(x) + C_2 y_2(x)$$, where $$C_1$$ and $$C_2$$ are arbitrary constants.

## How do you solve a non-homogeneous second-order linear differential equation?

To solve a non-homogeneous second-order linear differential equation $$a(x)y'' + b(x)y' + c(x)y = f(x)$$, one typically finds the general solution to the corresponding homogeneous equation $$a(x)y'' + b(x)y' + c(x)y = 0$$ and then finds a particular solution $$y_p(x)$$ to the non-homogeneous equation. The general solution to the non-homogeneous equation is then $$y(x) = y_h(x) + y_p(x)$$, where $$y_h(x)$$ is the general solution of the homogeneous equation.

## What methods can be used to find particular solutions to non-homogeneous second-order linear differential equations?

Several methods can be used to find particular solutions, including the method of undetermined coefficients, variation of parameters, and using Green's functions. The method of undetermined coefficients is useful when $$f(x)$$ is a simple function like a polynomial, exponential, sine, or cosine. Variation of parameters is a more general method that can be applied to a wider range of functions $$f(x)$$.

## What are characteristic equations, and how are they used in solving second-order linear differential equations with constant coefficients?

The characteristic equation is a quadratic equation derived from a second-order linear differential equation with constant coefficients. For the equation $$ay'' + by' + cy = 0$$, the characteristic equation is \( ar^2 +

Replies
18
Views
2K
Replies
4
Views
1K
Replies
6
Views
1K
Replies
1
Views
921
Replies
10
Views
1K
Replies
5
Views
1K
Replies
5
Views
1K
Replies
5
Views
3K
Replies
9
Views
1K
Replies
2
Views
986