Q3-T0101

[Victor Ivrii]

Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$
t^2y'' - t(t + 2)y' + (t + 2)y = 0.
$$

[Darren Zhang]

Divide both side by $t^2$, $$y''-\frac{t+2}{t}y'+\frac{t+2}{t^2}=0$$
The Wronskian of two solutions can be calculated by the formula,
$$c(exp[-\int \frac{t+2}{t}dt])$$
Then we can get the solution
$$c*(e^t*t^2)$$

[Mark Buchanan]

Divide everything by $t^2$ to get $y''$ by itself.
$$y'' - {t+2\over t}y' + {t+2\over t^2}y = 0$$

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 $- {t+2\over t}$ in this case.  Now we solve the integral:
$$ce^{\int{t+2\over t}dt}$$

Using the substitution $u = t+2$ and $du =dt$ we get
$$ ce^{\int{u\over u-2}du}$$

Which we can split up using partial fraction decomposition giving us
$$ce^{\int({2\over u-2}+1)du} = ce^{2\int({1\over u-2})du +\int{du}} = ce^{2ln(u-2)+u+C} = ce^{C}e^{2}e^{ln(t^2)}e^{t} = ce^{C}e^{2}t^2e^{t}$$

But $ce^Ce^2$ is just some constant, so we can subsume it into just $c$.  Simplifying this, we get that the Wronskian is:
$$W = {ct^2e^t}$$

[Meng Wu]

First, we divide both sides of the equation by $t^2$:
$$y''+{t+2\over t}y'+{t+2\over t^2}y=0$$
Now the given second-order differential equation has the form:
$$L[y]=y''+p(t)y'+q(t)y=0$$
Noting if we let $p(t)={t+2\over t}$ and $q(t)={t+2\over t^2}$, then $p(t)$ is continuous everywhere except at $t=0$, 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
$$\begin{align}W[y_1,y_2](t)&= cexp(-\int{p(t)dt})\\&=cexp(-\int{{t+2\over t}dt})\\&=ce^{t+2ln|t|}\\&=ct^2e^t\end{align} $$

[Victor Ivrii]

Meng Wu, there is no post solution after Mark.

Junjie, * is not the sign of multiplication (it is a sign of the different operation, called convolution)