Suppose i have the following causal and linear system
$$ ty''(t) +ty(t)=w'(t) $$
and i want to show that it is not time-invariant. I know that there are easier ways to show that it is not time-invariant, but suppose i want to do this by first finding the output of the input $w(t)=\theta(t)$. Setting $w(t) = \theta(t)$ and then solving the differentialequation by Laplacetransformation gives me the following output
$y_1(t)=-cos(t)\theta(t)+Csin(t)\theta(t)$
where C is a constant. Now if i take the input $w(t)=\theta(t-1)$ for example and plug it into the differential equation i get
$$ ty''(t) +ty(t)=\delta(t-1) $$
where $\delta(t)$ is the dirac delta function.
If i now again solve the differential equation by Laplacetransformation i get the output
$$y_2(t)=sin(t-1)\theta(t-1)-sin(t)\theta(t)+Csin(t)\theta(t)$$
and since $$y_2(t)\ne y_1(t-1)$$ for all t, the system cannot be time-invariant.
Is this valid way to show that the system is not time-invariant? Basically what i'm a bit uncertain about is whether the input $w(t)=\theta(t-1)$ should lead to the differential equation $$ ty''(t) +ty(t)=\delta(t-1) $$ OR $$ (t-1)y''(t-1) +(t-1)y(t-1)=\delta(t-1) $$
My intution says that it should lead to $$ ty''(t) +ty(t)=\delta(t-1) $$ but im not 100% sure.