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Question
Solve the following differential equation:
`"dy"/"dx" = (1 + "y")^2/(1 + "x")^2`
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Solution
`"dy"/"dx" = (1 + "y")^2/(1 + "x")^2`
∴ `1/(1 + "y"^2) "dy" = 1/(1 + "x"^2) "dx"`
Integrating both sides, we get
`int 1/(1 + "y"^2) "dx" = int1/(1 + "x"^2) "dx"`
∴ tan-1 y = tan-1 x + c
This is the general solution.
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