Question

solve the initial value problem:

dy/dx= (1+y^2)tanx , y(0) = square root of 3

Answer 1

You can solve it by separating variables: \(\frac{dy}{dx}=(1+y^2)\tan{x} \implies \frac{dy}{1+y^2}=\tan{x}dx\). By integrating both sides we get:
\(\int\frac{dy}{1+y^2}=\int\tan{x}dx \implies \tan^{-1}(y)=-\log(\cos{x})+c \implies y=\tan(-\log(\cos{x})+c).\)

Answer 2

Which is the same as \(y=-\tan(\log(\cos{x})-c)\).

Answer 3

Oh I forgot to substitute the initial value! Here:
\(\sqrt{3}=-\tan(\log(1)-c)=\tan(c) \implies c=\frac{\pi}{3}\). Hence \(y=-\tan(\log(\cos{x})+\frac{\pi}{3}).\)