Question

xy' + x^3 . lnx .y = 0

Answer 1

Explicação passo a passo:

Vamos encontrar a solução dessa equação:

$xy' + x^{3} .lnx.y = 0\\xy' = -x^{3}.lny\\ y' = -x^{2} .lnx.y$

$\frac{y'}{y} = -x^{2}.lnx\\\\\frac{dy}{dx}.\frac{1}{y} = -x^{2} .lnx\\\\\frac{1}{y} . dy = (-x^{2} .lnx)dx\\\\\int\limits^._. {\frac{1}{y}.dy } \, = \int\limits^._. {-x^{2}.lnx } \,dx \\\\lny = \int\limits^._. {-x^{2}.lnx } \,dx$

Vamos calcular a integral da direita utilizando a integração por partes:

$\int\limits^._. {-x^{2}.lnx } \,dx = -(\frac{x^{3} }{3} .lnx - \int\limits^._. {\frac{x^{3} }{3}.\frac{1}{x} } \, dx )\\\\\int\limits^._. {-x^{2}.lnx } \,dx = -\frac{x^{3} }{3} .lnx + \frac{1}{3} .\int\limits^._. {x^{2} } dx \\\\\int\limits^._. {-x^{2}.lnx } \,dx = -\frac{x^{3} }{3} .lnx + \frac{1}{3}.\frac{x^{3} }{3} \\\\\int\limits^._. {-x^{2}.lnx } \,dx = -\frac{x^{3} }{3} .lnx + \frac{x^{3} }{9} \\\\Portanto:\\\\ lny = -\frac{x^{3} }{3} .lnx + \frac{x^{3} }{9}$

$e^{lny} = e^{-\frac{x^{3} }{3} .lnx + \frac{x^{3} }{9}} \\\\y = e^{-\frac{x^{3} }{3} .lnx + \frac{x^{3} }{9}}$

Espero ter ajudado :)