- #1

Math100

- 773

- 219

- 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) 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) ##.