MAT244-2013S > Term Test 1

TT1--Problem 2

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Alexander Jankowski:
(a) We first re-write the differential equation as

\begin{equation*}
y'' - \frac{t\cos{t}}{\cos{t} + t\sin{t}}y' + \frac{\cos{t}}{\cos{t} + t\sin{t}} = 0.
\end{equation*}

Then, according to Abel's identity,

\begin{equation*}
W(y_1,y_2)(t) = c \exp{\left(-\int{\left(\frac{-t\cos{t}}{\cos{t} + t\sin{t}}\right)dt}\right)}.
\end{equation*}

Let $u = \cos{t} + t\sin{t}$ so that $du = t\cos{t}dt$. Then,

\begin{equation*}
W(y_1,y_2)(t) = c \exp{\int{\frac{du}{u}}} = c \exp {\ln{|t\sin{t} + \cos{t}|}} = c|t\sin{t} + \cos{t}|.
\end{equation*}

Using $W(0) = 1$ yields

\begin{equation*}
1 = C|0\cdot\sin{0} + \cos{0}| \Longleftrightarrow C = 1.
\end{equation*}

Therefore, the desired Wronskian is

\begin{equation*}
W(y_1,y_2)(t) = |t\sin{t} + \cos{t}|.
\end{equation*}

I do not yet know how to solve (b) without guessing, as applying reduction of order yields a rather unpleasant integral.

Victor Ivrii:

--- Quote from: Alexander Jankowski on February 14, 2013, 08:52:15 AM ---I do not yet know how to solve (b) without guessing, as applying reduction of order yields a rather unpleasant integral.

--- End quote ---

You don't need absolute value as a priory everything works on intervals where leading coefficient $\cos(t)+t\sin(t)$ does not vanish.

Also look at  J. Y. Yook' solution: knowing $W(y_1,y_2)$ and $y_1$ you get $1$-st order ODE for $y_2$ which could be solved. Alternatively (it is equivalent albeit more lengthy solution) you look $y_2=zy_1$ and you get $1$-st order ODE for $z'$.

Rudolf-Harri Oberg:
Part B, Alternative solution.
Assume the second solution is of the form $y_2=z(t)t$. Plugging that into the ODE, we get that
$tz''+(2+t \frac{-t \cos (t)}{\cos(t)+t \sin(t)})z'=0$ (look at reduction of order at pg 171, formula 30). This is a first order ODE for z'. So, plug for example $u=z'$. This equation is separable and simplifies to

$$tu'=(\frac{t^2 \cos(t)}{\cos(t)+t \sin(t)}-2)u$$
$$\frac{1}{u}du=\frac{1}{t}(\frac{t^2 \cos(t)}{\cos(t)+t \sin(t)}-2)dt$$.
Now "just" integrate and you get
$$\ln u= \ln(t\sin(t)+\cos(t))-2\ln (t) + \ln C_1$$ or
$$z'=u=\frac{t\sin(t)+\cos(t)}{t^2}$$ (forget the constant)
Now we have to integrate once again to get
$z=\frac{-\cos(t)}{t}+C$; so in conclusion we get that $y_2=zy_1=\frac{-\cos(t)}{t}t+Ct=-cos(t)+Ct$

We need $y_2(\frac{\pi}{2})=0$ which yields C=0. So, $y_2=-\cos(t)$ is the function we are looking for. Also note that 
$W(y_1,y_2)(\frac{\pi}{2})=-\cos(\frac{\pi}{2})+\frac{\pi}{2}\sin (\frac{\pi}{2})=\frac{\pi}{2}$, so the Wronskian condition is also satisfied.

Comment: I went for this solution idea during my test. Unfortunately, given half a page of room, exam pressure and those nasty integrals, I could not get to the final solution.

Victor Ivrii:
You cannot say  a priory that $C_1=0$ as we know $W(y_1,y_2)$ not up to a constant but exactly

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