- #1

Math100

- 779

- 220

- Homework Statement
- Let ## n\geq 1 ## be an integer and let ## F(y, z)=y^{n-1}\sqrt{y^2+z^2} ##.

a) Calculate ## G(y, z)=z\frac{\partial}{\partial z}F(y, z)-F(y, z) ##, and solve the equation ## G(y, z)=C ## for ## z ##, where ## C ## is a constant and ## y^{2n}\geq C^2 ##.

b) Let ## C>0, a>0, A>0 ## be constants, with ## na<\pi/2 ##. Show that the solution ## y(x) ## on the interval ## -a\leq x\leq a ## of the first-order differential equation ## C^{2}(\frac{dy}{dx})^2=y^2(y^{2n}-C^2), y(a)=A, y'(0)=0, y^{2n}\geq C^2 ##, is ## y(x)=A(\frac{cos(na)}{cos(nx)})^{1/n} ##. You may use the integral ## \int\frac{dy}{y(y^{2n}-C^2)^{1/2}}=\frac{1}{nc}arccos(\frac{C}{y^{n}})+constant ##, where ## n>0, C>0 ## are constants.

- Relevant Equations
- None.

a) Observe that ## \frac{\partial}{\partial z}F(y, z)=y^{n-1}\cdot \frac{2z}{2\sqrt{y^2+z^2}}=\frac{zy^{n-1}}{\sqrt{y^2+z^2}} ##.

This means ## G(y, z)=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-y^{n-1}\cdot \sqrt{y^2+z^2}=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-\frac{y^{n-1}(y^2+z^2)}{\sqrt{y^2+z^2}}=\frac{-y^{n+1}}{\sqrt{y^2+z^2}} ##.

Let ## G(y, z)=C ##.

Then ## \frac{-y^{n+1}}{\sqrt{y^2+z^2}}=C\implies \sqrt{y^2+z^2}=\frac{-y^{n+1}}{C} ##,

so ## y^2+z^2=\frac{y^{2n+2}}{C^2}\implies z^2=\frac{y^{2n+2}}{C^2}-y^2 ##.

Thus ## z=\pm\sqrt{\frac{y^{2n+2}-C^2y^2}{C^2}}=\pm\sqrt{\frac{y^2(y^{2n}-C^2)}{C^2}} ##.

Therefore, ## z=\pm\frac{y\cdot \sqrt{y^{2n}-C^2}}{C} ##.

b) Consider the first-order differential equation ## C^2(\frac{dy}{dx})^2=y^2(y^{2n}-C^2), y(a)=A, y'(0)=0, y^{2n}\geq C^2 ##.

Then ## C^2(\frac{dy}{dx})^2=y^2(y^{2n}-C^2) ##

## \frac{C^2(\frac{dy}{dx})^2}{y^2(y^{2n}-C^2)}=1 ##

## \frac{(\frac{dy}{dx})^2}{y^2(y^{2n}-C^2)}=\frac{1}{C^2} ##

## \frac{dy}{dx}=\pm\sqrt{\frac{1}{C^2}\cdot [y^2(y^{2n}-C^2)]} ##

## \frac{dy}{dx}=\pm\sqrt{\frac{1}{C^2}}\cdot \sqrt{y^2(y^{2n}-C^2)} ##

## \int\frac{dy}{\sqrt{y^2(y^{2n}-C^2)}}=\int\sqrt{\frac{1}{C^2}}dx ## since ## C>0 ##

## \int\frac{dy}{\sqrt{y^2}\cdot \sqrt{y^{2n}-C^2}}=\int\frac{\sqrt{C^2}}{C^2}dx ##

## \int\frac{dy}{y\cdot \sqrt{y^{2n}-C^2}}=\int\frac{\sqrt{C^2}}{C^2}dx ##.

Now we use the integral ## \int\frac{dy}{y(y^{2n}-C^2)^{1/2}}=\frac{1}{nC}arccos(\frac{C}{y^{n}})+constant ## to get

## \frac{1}{nC}arccos(\frac{C}{y^{n}})+constant=\frac{1}{C}x ##.

Let ## K=constant ##.

Then ## arccos(\frac{C}{y^{n}})=nC(\frac{1}{c}x-K)\implies \frac{C}{y^{n}}=cos(nx-nCK)\implies y^{n}=\frac{C}{cos(nx-nCK)} ##,

so ## y=(\frac{C}{cos(nx-nCK)})^{1/n}=(C\cdot sec(nx-nCK))^{1/n} ##.

Applying the initial condition of ## y(a)=A ## produces ## A=(\frac{C}{cos(na-nCK)})^{1/n}\implies A^{n}=\frac{C}{cos(na-nCK)} ##,

so ## C=A^{n}\cdot cos(na-nCK) ##.

Note that ## y'=\frac{1}{n}(C\cdot sec(nx-nCK))^{\frac{1-n}{n}}\cdot C\cdot sec(nx-nCK)\cdot tan(nx-nCK) ##.

Now we have ## y'(0)=0\implies 0=\frac{1}{n}(C\cdot sec(-nCk))^{\frac{1-n}{n}}\cdot c\cdot sec(-nCk)\cdot tan(-nCK) ##.

Thus ## 0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1-n}{n}}\cdot C\cdot sec(nCK)\cdot tan(nCK) ## (since ## sec(x) ## is an even function)

## 0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1}{n}-1}(C\cdot sec(nCK))\cdot tan(nCK) ##

## 0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1}{n}}\cdot tan(nCK) ## for ## n\neq 0, C>0 ##.

After simplifying the equation, we have ## tan(nCK)=0\implies K=0 ##.

Hence, ## C=A^{n}\cdot cos(na) ## and ## y=(A^{n}\cdot cos(na)\cdot sec(nx))^{1/n}=A(cos(na)\cdot \frac{1}{cos(nx)})^{1/n} ##.

Therefore, we have shown that the solution is ## y(x)=A(\frac{cos(na)}{cos(nx)})^{1/n} ##.

This means ## G(y, z)=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-y^{n-1}\cdot \sqrt{y^2+z^2}=\frac{z^2\cdot y^{n-1}}{\sqrt{y^2+z^2}}-\frac{y^{n-1}(y^2+z^2)}{\sqrt{y^2+z^2}}=\frac{-y^{n+1}}{\sqrt{y^2+z^2}} ##.

Let ## G(y, z)=C ##.

Then ## \frac{-y^{n+1}}{\sqrt{y^2+z^2}}=C\implies \sqrt{y^2+z^2}=\frac{-y^{n+1}}{C} ##,

so ## y^2+z^2=\frac{y^{2n+2}}{C^2}\implies z^2=\frac{y^{2n+2}}{C^2}-y^2 ##.

Thus ## z=\pm\sqrt{\frac{y^{2n+2}-C^2y^2}{C^2}}=\pm\sqrt{\frac{y^2(y^{2n}-C^2)}{C^2}} ##.

Therefore, ## z=\pm\frac{y\cdot \sqrt{y^{2n}-C^2}}{C} ##.

b) Consider the first-order differential equation ## C^2(\frac{dy}{dx})^2=y^2(y^{2n}-C^2), y(a)=A, y'(0)=0, y^{2n}\geq C^2 ##.

Then ## C^2(\frac{dy}{dx})^2=y^2(y^{2n}-C^2) ##

## \frac{C^2(\frac{dy}{dx})^2}{y^2(y^{2n}-C^2)}=1 ##

## \frac{(\frac{dy}{dx})^2}{y^2(y^{2n}-C^2)}=\frac{1}{C^2} ##

## \frac{dy}{dx}=\pm\sqrt{\frac{1}{C^2}\cdot [y^2(y^{2n}-C^2)]} ##

## \frac{dy}{dx}=\pm\sqrt{\frac{1}{C^2}}\cdot \sqrt{y^2(y^{2n}-C^2)} ##

## \int\frac{dy}{\sqrt{y^2(y^{2n}-C^2)}}=\int\sqrt{\frac{1}{C^2}}dx ## since ## C>0 ##

## \int\frac{dy}{\sqrt{y^2}\cdot \sqrt{y^{2n}-C^2}}=\int\frac{\sqrt{C^2}}{C^2}dx ##

## \int\frac{dy}{y\cdot \sqrt{y^{2n}-C^2}}=\int\frac{\sqrt{C^2}}{C^2}dx ##.

Now we use the integral ## \int\frac{dy}{y(y^{2n}-C^2)^{1/2}}=\frac{1}{nC}arccos(\frac{C}{y^{n}})+constant ## to get

## \frac{1}{nC}arccos(\frac{C}{y^{n}})+constant=\frac{1}{C}x ##.

Let ## K=constant ##.

Then ## arccos(\frac{C}{y^{n}})=nC(\frac{1}{c}x-K)\implies \frac{C}{y^{n}}=cos(nx-nCK)\implies y^{n}=\frac{C}{cos(nx-nCK)} ##,

so ## y=(\frac{C}{cos(nx-nCK)})^{1/n}=(C\cdot sec(nx-nCK))^{1/n} ##.

Applying the initial condition of ## y(a)=A ## produces ## A=(\frac{C}{cos(na-nCK)})^{1/n}\implies A^{n}=\frac{C}{cos(na-nCK)} ##,

so ## C=A^{n}\cdot cos(na-nCK) ##.

Note that ## y'=\frac{1}{n}(C\cdot sec(nx-nCK))^{\frac{1-n}{n}}\cdot C\cdot sec(nx-nCK)\cdot tan(nx-nCK) ##.

Now we have ## y'(0)=0\implies 0=\frac{1}{n}(C\cdot sec(-nCk))^{\frac{1-n}{n}}\cdot c\cdot sec(-nCk)\cdot tan(-nCK) ##.

Thus ## 0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1-n}{n}}\cdot C\cdot sec(nCK)\cdot tan(nCK) ## (since ## sec(x) ## is an even function)

## 0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1}{n}-1}(C\cdot sec(nCK))\cdot tan(nCK) ##

## 0=\frac{1}{n}(C\cdot sec(nCK))^{\frac{1}{n}}\cdot tan(nCK) ## for ## n\neq 0, C>0 ##.

After simplifying the equation, we have ## tan(nCK)=0\implies K=0 ##.

Hence, ## C=A^{n}\cdot cos(na) ## and ## y=(A^{n}\cdot cos(na)\cdot sec(nx))^{1/n}=A(cos(na)\cdot \frac{1}{cos(nx)})^{1/n} ##.

Therefore, we have shown that the solution is ## y(x)=A(\frac{cos(na)}{cos(nx)})^{1/n} ##.

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