$t^2y" + ty' + y = 0$
Let $x = lnt$, $t >0$, then $\frac{\partial x}{\partial t} = \frac{\partial }{\partial t}(lnt) = \frac{1}{t}$
Then we have $\frac{\partial y}{\partial t} = \frac{\partial y}{\partial x} \frac{1}{t}$
$\frac{\partial^2 y}{\partial t^2} = \frac{\partial }{\partial dt}(\frac{\partial y}{\partial x} \frac{1}{t}) = \frac{1}{t^2}(\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x}) $
When we substitute these to the original equation, we have :
$t^2 (\frac{1}{t^2}(\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x})) + t\frac{\partial y}{\partial x} \frac{1}{t} + y = 0$
$\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x} + \frac{\partial y}{\partial x} + y = 0$
$\frac{\partial^2 y}{\partial x} + y = 0$
Then we have a homogeneous equation, $y''+ y = 0$
$r^2 = -1$, $r = \pm i$, since $\lambda = 0$, $\mu = 1$
Then the general solution of this differential equation $y(x) = c_1 cos(x) + c_2 sin(x)$
And then substitute $x = lnt$, we get $y(t) = c_1 cos(lnt) + c_2 sin(lnt)$
