How do you find the general solution to #dy/dx=1/sec^2y#?

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Answer 1 · 2016-07-24T19:11:48

# y = tan^(-1) (x + C)#

we can separate it

#dy/dx=1/sec^2y#

#sec^2y dy/dx=1#

#int \ sec^2y dy/dx \ dx =int \ dx#

#int \d/dx( tan y) \ dx =int \ dx#

# tan y =x + C#

# y = tan^(-1) (x + C)#