MAT244-2013S > Ch 3

Bonus problem for week 3a

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Victor Ivrii:
(a). Reduce to the first order equation by substitution $y'=z$ and then find the general solutions (solutions using characteristic roots will not be considered)

\begin{equation}
y'' -\omega^2 y=0
\end{equation}


(b)  Also solve initial value problem
$y(0)=1$, $y'(0)=0$.

Brian Bi:
What's the difference between the 3a and 3b questions?

Changyu Li:
a)

$$
y'' - w^2 y = 0 \\
z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\
\frac{dz}{dy} z = w^2 y \\
\frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\
$$
\begin{equation}
\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1} \\
\end{equation}
factor out $w^2$ from $C_1$
$$
\frac{dy}{dx} = w \sqrt{y^2 + C_1} \\
\frac{dy}{ \sqrt{y^2 + C_1}} = w dx \\
$$
use trig substitution
$$
y = \sqrt{C_1} \tan u, \; dy=\sqrt{C_1} \sec^2 u\;du \\
\int \frac{\sqrt{C_1} \sec^2 u \; du}{\sqrt{C_1 \left( \tan^2 u + 1 \right)}} = \int wdx \\
\int \frac{\sqrt{C_1} \sec^2 u \; du}{\sqrt{C_1 \sec^2 u }} = \int wdx \\
\int \sec u \; du = \int wdx \\
\ln\left( \tan u + \sec\; u \right) = wx + C_2

$$
note $\sec \theta = \sqrt{1 + \tan^2 \theta}$ and $u = \arctan \frac{y}{\sqrt{C_1}}$
$$
\ln\left( \sqrt{y^2 + C_1 } + y \right) = wx + C_2 \\
\sqrt{y^2 + C_1 }  = e^{wx + C_2} - y \\
y^2 + C_1  = \left( e^{wx + C_2} - y\right)^2 \\
$$
expand and simplify
$$
y = \left( e^{2\left(wx+C_2\right)} - C_1 \right) e^{-\left(wx+C_2\right)} \frac{1}{2}
$$

b)
insert IC to (2) to get $C_1 = -w^2$
$$
\frac{dy}{dx} = w \sqrt{y^2-1} \\
\frac{1}{\sqrt{y^2-1}}  dy = w dx \\
\ln \left( \sqrt{ y^2 -1} + y \right) = wx + C_2 \\$$

insert IC to get $C_2 = \ln\left( i \right)$.
simplify using euler's identity
$$
e^{i\frac{\pi}{2}} = i \\
i\frac{\pi}{2} = \ln i \\
$$

$$
y = \left( e^{2\left(wx+i\frac{\pi}{2} \right)} + 1\right) e^{-\left(wx+i\frac{\pi}{2} \right)} \frac{1}{2}
$$

Victor Ivrii:
1) After $\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1}$ there is no point to "insert" something--at least not in the general solution. One can perfectly integrate $\frac{dy} {\sqrt{w^2 y^2 + w^2 A}}$ (better to avoid indices).

2) After $\ln \bigl(y+\sqrt{y^2 + A}\bigr)= \ldots$ derived one needs to find $y=\ldots$

Changyu Li:
fixed.

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