MAT244-2018S > Quiz-3

Q3-T0301

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**Victor Ivrii**:

Find the Wronskian of two solutions of the given differential equation without solving the equation.

$$

\cos (t)y'' + \sin (t)y' - ty = 0.

$$

**Mark Buchanan**:

Divide everything by $cos(t)$ to get $y''$ by itself.

$$y'' + {sin(t)\over cos(t)}y' - {t\over cos(t)}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(t)dt}$ where c is a constant and $p(t)$ is $sin(t)\over cos(t)$ in this case. Now we solve the integral:

$$ce^{-\int{sin(t)\over cos(t)}dt}$$

Using the substitution $u=cos(t)$ and $du=-sin(t)dt$ we get

$$ce^{\int{1\over u}du}=ce^{ln(u)+C}=ce^{ln(cos(t)+C}=ce^Ccos(t)$$

But $ce^Ce^2$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c(cos(t))}$$

**Meng Wu**:

First, we divide both sides of the equation by $cos(t)$:

$$y''+tan(t)y'-{t\over cos(t)}y=0$$

Now the given second-order differential equation has the form:

$$L[y]=y''+p(t)y'+q(t)y=0$$

Noting if we let $p(t)=tan(t)$ and $q(t)=-{t\over cos(t)}$, then $p(t)$ is continuous everywhere except at ${\pi\over 2}+k\pi$, where $k=0,1,2,\dots$ and $q(t)$ is also continuous everywhere except at $t=0$. $\\$

Therefore, by Abel's Theorem: the Wronskian $W[y_1,y_2](t)$ is given by

$$\begin{align}W[y_1,y_2](t)&= cexp(-\int{p(t)dt})\\&=cexp(-\int{tan(t)dt})\\&=ce^{ln|cos(t)|}\\&=ccos(t)\end{align}$$

**Victor Ivrii**:

use \cos \sin etc

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