How to you find the general solution of $dy/dx=(x^2+2)/(3y^2)$ ?

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Answer 1 · 2017-01-12T17:10:43

$y(x) = root(3)(x^3/3+2x+C$

This is a separable differential equation, so we can proceed by separating the variables and integrating:

$(dy)/(dx) = (x^2+2)/(3y^2)$

$3y^2dy = (x^2+2)dx$

$int 3y^2dy = int(x^2+2)dx$

$y^3 = x^3/3+2x+C$

$y(x) = root(3)(x^3/3+2x+C$

How to you find the general solution of dy/dx=(x^2+2)/(3y^2)dy/dx=(x^2+2)/(3y^2)?