### Author Topic: Bonus problem for week 3b  (Read 6718 times)

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Bonus problem for week 3b
« on: January 24, 2013, 09:56:47 AM »
(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)

y'' + \omega^2 y=0

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

#### Changyu Li

• Full Member
• Posts: 16
• Karma: 10
##### Re: Bonus problem for week 3b
« Reply #1 on: January 24, 2013, 01:07:24 PM »
$$y'' + w^2 y = 0 \\ z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\ \frac{dz}{dy} z = -w^2 y \\ \frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\$$

\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\

insert IC to get $C_1 = w^2$
$$\frac{dy}{dx} = w \sqrt{1 - y^2} \\ \frac{1}{\sqrt{1-y^2}} dy = w dx \\ wx = \arcsin y + C_2 \\$$
insert IC to get $C_2 = -\pi / 2$
$$y = \sin\left( wx + \pi/2 \right) \\ y = \cos\left( wx \right)$$

general solution to $\frac{dy}{dx}= w \sqrt{C_1-y^2}$ using trig substitution
$$\frac{dy}{\sqrt{C_1-y^2}} = w dx \\ wx = \frac{\arcsin\left( y / \sqrt{C_1} \right)}{\sqrt{C_1}} + C_2 \\ y = \sqrt{C_1} \sin \left( \sqrt{C_1}\left( wx + C_2 \right) \right)$$
« Last Edit: January 24, 2013, 03:04:10 PM by Changyu Li »

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Re: Bonus problem for week 3b
« Reply #2 on: January 24, 2013, 02:09:42 PM »
You found correctly solution for initial value problem. What about general solution (again $\frac{dy}{dx}= w \sqrt{A-y^2}$).

#### Brian Bi

• Moderator
• Full Member
• Posts: 31
• Karma: 13
##### Re: Bonus problem for week 3b
« Reply #3 on: January 24, 2013, 11:19:31 PM »

\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\

separate variables

\frac{1}{\sqrt{C_1 - \omega^2 y^2}} \, dy = \, dx

integrate LS by substituting $y = \frac{\sqrt{C_1}}{\omega}u$
\begin{align}
\int \frac{1}{\sqrt{C_1 - \omega^2 y^2}} \, dy
&= \int \frac{1}{\sqrt{C_1 - \omega^2 ((\sqrt{C_1}/\omega)u)^2}} \frac{\sqrt{C_1}}{\omega} \, du \\
&= \int \frac{1}{\sqrt{C_1(1 - u^2)}} \frac{\sqrt{C_1}}{\omega} \, du \\
&= \frac{1}{\omega} \arcsin u + C_2 \\
&= \frac{1}{\omega} \arcsin\left(\frac{\omega}{\sqrt{C_1}}y\right) + C_2
\end{align}
Integrate also the RS to obtain

\frac{1}{\omega} \arcsin\left(\frac{\omega}{\sqrt{C_1}}y\right) + C_2 = x

Isolate $y$ in terms of $x$:

y = \frac{\sqrt{C_1}}{\omega} \sin(\omega x - C_2)

Expand the sine term:

y = \frac{\sqrt{C_1}}{\omega} (\sin(\omega x) \cos C_2 - \sin(C_2) \cos(\omega x))

By choosing $C_1 = \omega^2$ and either $C_2 = 0$ or $C_2 = 3\pi/2$ respectively we obtain particular solutions $y_1 = \sin(\omega x)$ and $y_2 = \cos(\omega x)$. The Wronskian is
\begin{align}
W &= \left|\begin{array}{cc} \sin(\omega x) & \cos(\omega x) \\ \omega \cos(\omega x) & -\omega \sin( \omega x) \end{array}\right| \\
&= -\omega \sin^2(\omega x) - \omega \cos^2(\omega x) \\
&= -\omega \neq 0
\end{align}
so we conclude that the general solution to the linear homogeneous second-order ODE is $\{y = A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$
« Last Edit: January 25, 2013, 12:38:12 AM by Brian Bi »

#### Changyu Li

• Full Member
• Posts: 16
• Karma: 10
##### Re: Bonus problem for week 3b
« Reply #4 on: January 24, 2013, 11:38:01 PM »
$y = \{A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$
let me plot that for you

#### Brian Bi

• Moderator
• Full Member
• Posts: 31
• Karma: 13
##### Re: Bonus problem for week 3b
« Reply #5 on: January 25, 2013, 12:54:10 AM »
I'm not sure whether that image got corrupted on upload or whether you're just pointing out that I erred in putting $y$ outside the braces instead of inside where it should be. (I fixed that now, btw.)

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Re: Bonus problem for week 3b
« Reply #6 on: January 25, 2013, 01:06:59 AM »
So, one can rewrite solution in  either form

y= A \cos (\omega t) + B\sin (\omega t)
\label{eq-a}

or

y= C \cos (\omega (t-t_0))
\label{eq-b}

where (\ref{eq-a}) is what we usually get from linear equations point of view (Ch 3) and  (\ref{eq-b}) what we get from autonomous systems (not necessarily linear, and we will be more interested in non-linear--Chapter 9)

PS. Try to add an image as an attachment (preferred) rather than upload to an external website where it got borked beyond recovery. As we would like to keep class archives for a future generations, we don't want rely upon other websites