You need to assume that the general form of solution to differential equation is `y = x^r` .

Differentiating y with respect to x yields:

`y' = rx^(r-1)`

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

Substituting y'' and y' in equation yields:

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

You need to factor out `x^r` such that:

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

Dividing by `x^r` yields:

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

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

You need to find the roots of quadratic equation such that:

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

`r(r+1) + 2(r+1) = 0`

Factoring out r+1 yields:

`(r+1)(r+2) = 0`

You need to solve equations r + 1 = 0 and r + 2 = 0 such that:

`r + 1 = 0 =gt r_1 = -1`

`r + 2 = 0 =gt r_2 = -2`

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

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now