How Do You Solve the Homogeneous Equation for ty''-(t+1)y'+y=0?

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$$ty''-(t+1)y'+y=t^2$$

I know I have to use variation of parameters to solve this.
But I am stuck and cannot figure out how to get the homologous equation!
$$y''-(1+\frac{1}{t})y'+\frac{1}{t}*y=t$$

I don't know how to solve this homologous equation in this format.

Is it R^2+(1+1/t)R+1/t = 0 ?
How would I get my two solutions from this?

Thanks!

 

[[V]]

Whats wrong with the latex??

 

[LCKurtz]

$$ty''-(t+1)y'+y=t^2$$

I know I have to use variation of parameters to solve this.
But I am stuck and cannot figure out how to get the homologous equation!
$$y''-(1+\frac{1}{t})y'+\frac{1}{t}*y=t$$

I don't know how to solve this homologous equation in this format.

Is it R^2+(1+1/t)R+1/t = 0 ?
How would I get my two solutions from this?

Thanks!

Your homogeneous equation is

$$ty''-(t+1)y'+y=0$$

This equation has a regular singular point at t = 0 suggesting you look for series solutions of the form

$$y = \sum_{n=0}^{\infty}a_nt^{n+r}$$