Let $\mathcal{L}$ be the Laplace transform function, and let $\mathcal{L}\bigl(f(t)\bigr) = F(s)$.
One of the properties of the Laplace transform is that $\mathcal{L}\Bigl((-t)^nf(t)\Bigr) = F^{(n)}(s)$, i.e., the $n$-th derivative of the transform with respect to $s$. I was testing this property out a bit.
Given $ty'+y=0$, find $y(t)$. The solution is very obvious when you use integrating factors: $$(ty)'=0$$ $$y=\frac{C}t$$
Now, let's try solving this differential equation using a Laplace transform!
$$\mathcal{L}\bigl(ty'+y\bigr)=0 \tag{1}\label{1}$$
The transform of $f(t)=y'$ is $sF(s)-f_0$, so then... $$\mathcal{L}\big(ty'\big) = -\frac{d\left(sF(s)-f_0\right)}{ds}=-sF'(s)-F(s) \tag{2}$$
Returning back to $(1)$, we now have $-sF'(s)-F(s)+F(s)=0$
which in turn simplifies to $$F'(s)=0$$
so $$F(s)=C$$
and $$f(t)=C\cdot\delta(t)$$
Can someone explain to me why what I did in $(2)$ isn't necessarily correct, because that's the step I'm iffy on right now. Thank you!
The problem is that the Laplace transform of $\frac{1}{t}$ does not exist (see this question). So this DE cannot be solved using the Laplace transform.