1 Answer | Add Yours
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` .
We’ve answered 319,865 questions. We can answer yours, too.Ask a question