You should assume that all solutions to equation are of form `y= x^r =gt y' = r*x^(r-1) and 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) + 5x^r = 0`

`r(r-1)*x^r + 3r*x^r + 5x^r = 0 `

Factoring out `x^r` yields:

`x^r *(r(r-1) + 3r + 5) = 0`

Dividing by `x^r` yields:

`r(r-1) + 3r + 5 = 0`

Opening the brackets yields:

`r^2 - r + 3r + 5 = 0`

`r^2 + 2r + 5 = 0`

You should use quadratic formula to find the roots such that:

`r_(1,2) = (-2+-sqrt(4 - 20))/2 =gt r_(1,2) = (-2+-sqrt(-16))/2`

`r_(1,2) = (-2+-4i)/2 =gt r_(1,2) = (-1+-2i)`

**Hence, evaluating the general solution to differential equation yields `y = c_1*x^(-1)*cos(2lnx) + c_2*x^(-1)*sin(2lnx).` **

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