How do you find the general solution to $dy/dx=2yx+yx^2$ ?

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Answer 1 · 2016-07-24T17:40:55

$y = Ce^(x^2 + x^3 /3 )$

$dy/dx=2yx+yx^2$

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

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

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

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

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

$y = Ce^(x^2 + x^3 /3 )$

How do you find the general solution to dy/dx=2yx+yx^2dy/dx=2yx+yx^2?