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))#