Question

x^2y''-xy'=lnx

Answer 1

$x^{2}y'' - xy' = lnx$

$y' = p(x), y'' = p'(x)$

$x^{2}p' -xp = lnx$

$p = uv, p' = u'v + uv'$

$x^{2}(u'v + uv') - xuv = lnx$

$v(x^{2}u' - xu) + x^{2}uv' = lnx$

$x^{2}u' - xu = 0, u=x$

$x^{3}v' = lnx$

$v = \int \frac{lnx}{x^{3}} dx = [x = e^{p}, dx = e^{p}dp, lnx = p] = \int pe^{-2p}dp = - \frac{1}{2}(pe^{-2p} -\int e^{-2p}dp)=$

$=- \frac{1}{2}(pe^{-2p} +\frac{1}{2}e^{-2p}) = [x=e^{p}] = -\frac{1}{2x^{2}}(lnx+\frac{1}{2} + c)$

$y' = p = uv = -\frac{1}{2x}(lnx+\frac{1}{2} + c)$

$y = \int -\frac{1}{2x}(lnx+\frac{1}{2} + c)dx = \int -\frac{lnx}{2x} dx - \frac{1}{4}ln|x|- \frac{c_1}{2} ln|x| + c_2=$

$- \frac{ln^{2}|x|}{4} - \frac{1}{4}ln|x|- \frac{c_1}{2} ln|x| + c_2$