MAT244-2013S > Ch 3

Bonus problem for week 3a

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Brian Bi:

--- Quote from: Changyu Li on January 24, 2013, 01:33:03 PM ---$$
y = \left( e^{2\left(wx+i\frac{\pi}{2} \right)} + 1\right) e^{-\left(wx+i\frac{\pi}{2} \right)} \frac{1}{2}
$$

--- End quote ---
Pretty sure this isn't right; the correct solution is $y = \frac{1}{2}\left( e^{\omega x} + e^{-\omega x}\right)$. How exactly did you calculate $C_1$ and $C_2$?

BTW, this problem can be solved analogously to the 3b problem by using a hyperbolic trig substitution instead of a circular trig substitution. The result is, unsurprisingly, $y = A \cosh \omega x + B \sinh \omega x$. This form is more convenient for the initial conditions given because here $y(0) = A$ and $y'(0) = \omega B$, whence $A = 1, B = 0$ so indeed $y = \cosh \omega x$.

Victor Ivrii:
After you got

--- Quote from: Changyu Li on January 24, 2013, 01:33:03 PM ---\begin{equation*}
\ln\left( \sqrt{y^2 + C_1} + y \right) = \omega x + D
\end{equation*}

--- End quote ---
which could be rewritten as either
\begin{equation}
\ln\left( \sqrt{y^2 - C^2} + y \right) = \omega x + D \iff  \ln\left( \sqrt{C^{-2}y^2 -1} + C^{-1}y \right) = \omega x + D_1
\label{eq-a}
\end{equation}
as $C_1=-C^2<0$  or
\begin{equation}
\ln\left( \sqrt{y^2 +C^2} + y \right) = \omega x + D \iff  \ln\left(\sqrt{C^{-2}y^2 +1} + C^{-1}y \right) = \omega x + D_1
\label{eq-b}
\end{equation}
as $C_1=C^2>0$, you can extract  $y= C\cosh(\omega x+D_1)$ and $y= C\sinh (\omega x+D_1)$ respectively. As $C=1$, $D=0$ we get $y_1=\cosh (\omega x)$ and  $y_2=\sinh (\omega x)$ which provide another basis in the space of solutions (in comparison with $\bigl\{e^{\omega x}, e^{-\omega x}\bigr\}$). The same is true for any choice of $C$ and $D_1$ (just different basis).

See remark to a "sister problem" 3b.

Changyu Li:
To solve for $C_1$ and $C_2$, I tried substituting $dy/dx = 0,\;y = 1$ into $\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1}$ and $x = 0,\;y = 1$ into $\ln \left( \sqrt{ y^2 -1} + y \right) = wx + C_2$. I'm not sure why it gives a different result.

Victor Ivrii:
OK, WTH is $\ln (\sqrt{y^2\mp 1}+y)= x$? It is equivalent to
$$\sqrt{y^2\mp 1}=e^{x}-y\implies y^2\mp 1=e^{2x}-2ye^{x}+y^2\implies y= \frac{e^{x}\pm e^{-x}}{2}$$
which is exactly $\cosh(x)$ and $\sinh(x)$ respectively. So $\ln (\sqrt{y^2\mp 1}+y)$ are just inverse hyperbolic $\cosh$ and $\sinh$.

Hyperbolic functions are very useful as they are in many senses similar to trigonometric functions.

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