### Author Topic: Q3-T0301  (Read 3143 times)

#### Victor Ivrii ##### Q3-T0301
« on: February 10, 2018, 05:16:49 PM »
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

• Jr. Member
•  • Posts: 10
• Karma: 4 ##### Re: Q3-T0301
« Reply #1 on: February 10, 2018, 06:05:49 PM »
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 ##### Re: Q3-T0301
« Reply #2 on: February 11, 2018, 10:06:57 AM »
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 ##### Re: Q3-T0301
« Reply #3 on: February 21, 2018, 07:48:37 AM »
use \cos \sin etc