Toronto Math Forum
MAT2442013S => MAT244 MathTests => Term Test 1 => Topic started by: Victor Ivrii on February 13, 2013, 10:38:31 PM

(a) Consider equation
\begin{equation*}
(\cos(t)+t\sin(t))y''t\cos(t)y'+y\cos(t)=0.
\end{equation*}
Find wronskian $W=W[y_1,y_2](t)$ of two solutions such that $W(0)=1$.
(b) Check that one of the solutions is $y_1(t)=t$. Find another solution $y_2$ such that $W[y_1,y_2](\pi/2)=\pi/2$
and $y_2(\pi/2)=0$.

part a)

Part (b).
Can I please get bonus marks? :'( I couldn't do it under limited time and exam pressure. I don't know how how I couldn't see things in the exam center room. Now things work out....

part a)

See my comments to Problem 1. Waiting for typed solution.

(a) We first rewrite 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}}} = ct\sin{t} + \cos{t}.
\end{equation*}
Using $W(0) = 1$ yields
\begin{equation*}
1 = C0\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.

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

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.

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