How do you find the general solution to $(x^2+1)y'=xy$ ?

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Answer 1 · 2016-07-24T16:13:01

$y =sqrt ( C(x^2+1))$

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

separate it

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

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

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

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

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

$y =sqrt ( C(x^2+1))$

How do you find the general solution to (x^2+1)y'=xy(x^2+1)y'=xy?