You need to remember that you may assume that the general form of all solutions to differential equation is `y = x^r` .

Differentiating with respect to x yields:

`y' = r*x^(r-1)`

`y'' = r(r-1)*x^(r-2)`

You need to substitute `r(r-1)*x^(r-2)` for y'' and `r*x^(r-1)` for y' in equation such that:

`x^2*r(r-1)*x^(r-2) - 3x*r*x^(r-1) + 4x^r = 0`

`r(r-1)*x^r - 3r*x^r+ 4 x^r = 0`

Factoring out `x^r` yields:

`x^r(r(r - 1) - 3r + 4) = 0`

You need to divide by `x^r` such that:

`r^2 -r - 3r + 4 = 0`

`r^2 -4r +4 = 0 =gt (r - 2)^2 = 0 =gt r_1 = r_2 = 2`

**Hence, evaluating the general solution to differential equation yields `y = c_1*x^2 + c_2*x^2*lnx` .**

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