I need to find the general solution of the differential equation:
(can't get the next bit to work in LaTeX so sorry about this)
x = integral of 1/(1+y^2) dy
= inverse tan(y) + k
x-k = inverse tan (y)
y = tan(x-k)
but the answer should be y = tan(x + Pi/4 -2)
Any ideas where I'm going wrong? Thanks
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- Thread Starter
- 01-02-2010 19:53
- 01-02-2010 20:47
Your answer isn't wrong.
y = tan(x-k) is the general solution of the differential equation.
y = tan(x + Pi/4 -2) is the general solution subject to a condition i.e. when x = ... , y = ...