MAT244-2013S > Ch 3

Bonus problem for week 3b

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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
$y(0)=1$, $y'(0)=0$.

Changyu Li:
$$
y'' + w^2 y = 0 \\
z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\
\frac{dz}{dy} z = -w^2 y \\
\frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\
$$
\begin{equation}
\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\
\end{equation}
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)
 
$$

Victor Ivrii:
You found correctly solution for initial value problem. What about general solution (again $\frac{dy}{dx}= w \sqrt{A-y^2}$).

Brian Bi:

--- Quote from: Changyu Li on January 24, 2013, 01:07:24 PM ---\begin{equation}
\frac{dy}{dx} = \sqrt{-w^2 y^2 + C_1} \\
\end{equation}

--- End quote ---
separate variables
\begin{equation}
\frac{1}{\sqrt{C_1 - \omega^2 y^2}} \, dy = \, dx
\end{equation}
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
\begin{equation}
\frac{1}{\omega} \arcsin\left(\frac{\omega}{\sqrt{C_1}}y\right) + C_2 = x
\end{equation}
Isolate $y$ in terms of $x$:
\begin{equation}
y = \frac{\sqrt{C_1}}{\omega} \sin(\omega x - C_2)
\end{equation}
Expand the sine term:
\begin{equation}
y = \frac{\sqrt{C_1}}{\omega} (\sin(\omega x) \cos C_2 - \sin(C_2) \cos(\omega x))
\end{equation}
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 $\{y = A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$

Changyu Li:
$y = \{A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$
let me plot that for you

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