How you solve this? $x^2y'+x^2y^2-3xy+3=0$ with solution $y_1=3/x$

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Answer 1 · 2018-04-11T20:50:18

$y = (C_1+3C_2x^2)/(C_1 x+ C_2 x^3)$

Making the substitution

$y = (xi')/xi$ we have after substitution

$x^2xi''-3x xi'+3xi = 0$

this is linear differential equation with solution

$xi(x) = C_1 x+ C_2 x^3$ then

$y = (C_1+3C_2x^2)/(C_1 x+ C_2 x^3)$

How you solve this? x^2y'+x^2y^2-3xy+3=0x^2y'+x^2y^2-3xy+3=0 with solution y_1=3/xy_1=3/x