Verify that the given functions are solutions to the corresponding homogenous equation $t^2y'' - 2y = 3t^2 - 1 | y_1(t) = t^2, y_2(t) = t^{-1}$ "verify that the given functions y1 and y2 satisfy the corresponding
homogeneous equation"
$t^2y'' - 2y = 3t^2 - 1$
$y1(t) = t^2$
$y2(t) = t^{-1}$
However when I attempt to do this I find that they do not satisfy, what am I doing wrong here? I start by differentiating y1
$y1(t) = t^2$
$y1'(t) = 2t$
$y1''(t) = 2$
Use these to plug into the original equation
$t^2y'' - 2y = 3t^2 - 1$
$t^2(2) - 2(t^2) = 3t^2 - 1$
$0 = 3t^2 - 1$
I get similar results doing the same thing with y2(t).
 A: To show that the given functions, 
$$ y_1(t)=t^2$$
$$y_2(t)=\frac{1}{t}$$
Not that these two solutions are linearly independent of each other.
Are solution to the homogeneous equation.
Given that the second order non-homogeneous differential equation is
$$t^2\frac{d^2y}{dt^2}-2y=3t^2-1$$
To make it into homogeneous form which is 
$$a_0(x)\frac{d^2y}{dx^2}+a_1(x)\frac{dy}{dx}+a_2(x)y=0$$
The homogeneous form of the differential equation is
$$t^2\frac{d^2y}{dt^2}-2y=0$$
To show that $y_1(t)$ is a solution to the differential equation, we just need to differentiate and plug into the homogeneous differential equation.
$$y=t^2$$
$$\frac{dy}{dt}=2t$$
$$\frac{d^2y}{dt^2}=2$$
$$t^2(2)-2(t^2)=0$$
Which is true
To show that $y_2(t)$ is a solution we then have
$$y=\frac{1}{t}$$
$$\frac{dy}{dt}=-\frac{1}{t^2}$$
$$\frac{dy}{dt}=\frac{2}{t^3}$$
And plug it in
$$t^2(\frac{2}{t^3})-2(\frac{1}{t})=0$$
$$\frac{2}{t}-\frac{2}{t}=0$$
Which is also true.
The above is what you need to do to show that they are solutions to the homogeneous form of the differential equation.
Sometimes, $y_1(t)$ & $y_2(t)$ are called complementary solutions to the non-homogeneous differential equation. 
Comments
In your question you are asked to solved homogeneous differential equation. However, you are solving the non-homogeneous differential equation and obviously it won't work.
