How to solve $ t^2 y''(t) -(t+2) t y'(t) +(t+2) y(t) = 0 $ with varying coefficients? This second order linear ODE, with varying coefficients,  is not of type Euler, and can't be converted to constant coefficient ODE using standard transformation.
I am not looking for series solution. I know I can solve this using series method if needed.
$$ 
t^2 y''(t) -(t+2) t y'(t) +(t+2) y(t) = 0
$$
Maple says this is solved using linear symmetries. I have not yet studied using symmetry transformation for second order ODE's and have not yet found easy introduction on this to follow. But will keep looking.
I am sure this is solved using transformation on the independent variable. But I do not know what this transformation is. Maple help page on this is below, but it is not clear to me.
I can solve Euler $t^2 y''+t y' + y=0$ and $y''+p(t) y'(t) + q(t) y=0$ where it is possible to convert it to constant coefficient using known transformation.
But this ODE is not one of these two types.
Any suggestions how to solve this?
Maple gives the solution $y(t) = c_2 t+ c_1 t e^t  $
Reference https://fr.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/linear_sym
 A: Let $y(t)=t\, z(t)$ wich makes the equation to be
$$z''(t)-z'(t)=0$$ Use reduction of order to get $z'(t)$ and integrate again to get $z(t)$ and then $y(t)$;
A: First divide out by $t^2$:
$$y''(t)-\frac{t+2}{t}y'(t)+\frac{t+2}{t^2}y(t)=0$$
Guess a second order polynomial solution: $y(t)=at^2+bt+c$ :
$$2a-\frac{t+2}{t}(2at+b)+\frac{t+2}{t^2}(at^2+bt+c)=0$$
$$2a-2at\frac{t+2}{t}-b\frac{t+2}{t}+at^2\frac{t+2}{t^2}+bt\frac{t+2}{t^2}+c\frac{t+2}{t^2}$$
$$2a-2a(t+2)-b\frac{t+2}{t}+a(t+2)+b\frac{t+2}{t}+c\frac{t+2}{t^2}=0$$
$$-at^3+c(t+2)-bt(t+2)+bt(t+2)=0$$
It is clear that any choice of $b$ will solve the above equation. The rest is a polynomial and thus only has finitely many zeroes. Therefore, for the remaining to be $=0~\forall t$, we must have that $a=c=0.$ So our first homogeneous solution is
$$y_1(t)=c_1t.$$ Since this is a linear second order ODE we can use Abel's Identity and see that
$$W(y_1,y_2)(t)=c_2\cdot \exp\left(-\int -\frac{t+2}{t}\mathrm{d}t\right)$$
Here $W$ is the Wronskian determinant:
$$W=\det(\mathbf{W})=\det\left(\begin{bmatrix}
y_{1} & y_{2}\\
{y_{1}} ' & {y_{2}} '
\end{bmatrix}\right)=y_1{y_2}'-y_2{y_1}'$$
Thus,
$$c_1(t{y_2}' -y_2)=c_2\cdot \exp\left(\int \frac{t+2}{t}\mathrm{d}t\right)$$
Some elementary integration shows that
$$\int\frac{t+2}{t}\mathrm{d}t=t+2\ln(t)+C$$
So
$$c_2\cdot \exp\left(\int\frac{t+2}{t}\mathrm{d}t\right)=C\cdot e^t t^2$$
Thus we now need only solve the ODE
$${y_2}'-\frac{1}{t}y_2=c_2\cdot t e^t $$
This can be solved using integrating factors. We can see that
$$y_2(t)=Ct+c_2 te^t$$
The $Ct$ is extraneous as it can be combined into $y_1$. So
$$y_2(t)=c_2 te^t$$
Thus finally,
$$y(t)=c_1 t+c_2 te^t.$$
A: $$t^2 y''(t) -(t+2) t y'(t) +(t+2) y(t) = 0$$
$y=t$ is an obvious solution:
$$t^2 \times 0-t(t+2)\times 1 +(t+2)\times t=0$$
Then use method of reduction of order:
$y=v(t)t$
