Solve #dy/dx = 1+ 1/y^2# ?

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Answer 1 · 2017-12-06T23:05:57

# y - arctan(y) = x + C #

We have:

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

Which is a First Order Separable Ordinary Differential Equation so we can rearrange and "separate the variables":

# dy/dx = (1+y^2)/y^2 #

# => int \ y^2/(1+y^2) \ dy = int \ dx #

We can manipulate the LHS integral:

# \ \ \ \ \ int \ (1+y^2-1)/(1+y^2) \ dy = int \ dx #

# :. int \ 1 - 1/(1+y^2) \ dy = int \ dx #

Which is now trivial to integrate giving us:

# y - arctan(y) = x + C #

Which is the general implicit solution.