What is the general solution of the differential equation? : $y' = x(1+y^2)$

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Answer 1 · 2018-02-07T18:00:49

$y = tan(1/2x^2 + C)$

We have:

$y' = x(1+y^2)$

This is a First Order Separable Ordinary Differential Equation. We can rewrite in the form:

$1/(1+y^2) dy/dx = x$

So we can "separate the variables" to get:

$int \ 1/(1+y^2) \ dy = int \ x \ dx$

Both integrals are standard calculus results , so integrating we get:

$arctany = 1/2x^2 + C$

Leading to the General Solution:

$y = tan(1/2x^2 + C)$

What is the general solution of the differential equation? : y' = x(1+y^2) y' = x(1+y^2)