Question #8b242

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Answer 1 · 2016-08-29T19:58:25

$y =1 - 3/(4x^3 -1)$

$y' = 4x^2 (y-1)^2, y(0) =1$

This is separable:

$1/(y-1)^2 y' = 4x^2$

$int 1/(y-1)^2 y' dx =int 4x^2 dx$

$int 1/(y-1)^2 dy =int 4x^2 dx$

$-1/(y-1) = (4x^3)/3 + C$

$-(y-1) = 1/((4x^3)/3 + C) = 3/(4x^3 + C)$

$y =1 - 3/(4x^3 + C)$

$y(1) = 0 = 1 - 3/(4 + C)$

$4 + C = 3, C = -1$

$y =1 - 3/(4x^3 -1)$