Originally Posted by

**Fallen186** I have the first order Initial Value Problem ODE of

$\displaystyle y' = 1 + 4y^2$ , $\displaystyle y = \frac{1}{2}*tan(2x+C)$ , $\displaystyle y(0) = 0$

I calculate that $\displaystyle y' = sec^2(2x+c)$

$\displaystyle

sec^2(2x+ C) = 1 + 4(\frac{1}{2}tan(2x+C))^2

$

$\displaystyle sec^2(2x+ C) = 1 + 4(\frac{1}{4}tan^2(2x+C))$

$\displaystyle sec^2(2x+ C) = 1 + tan^2(2x+C))$

*** trig identity$\displaystyle sec^2(u) = 1 + tan^2(u)$ ***

y(0) = 0 so

$\displaystyle sec^2(2x+C) = 1 + tan^2(2x+C))$

$\displaystyle sec^2(2(0) + C) = 1 + tan^2(2(0) + C)$

$\displaystyle sec^2(C) = 1 + tan^2(C)$

$\displaystyle \frac{1}{\frac{sin(C)}{cos(C)}} = 1 + \frac{sin(C)}{cos(C)}$

$\displaystyle \frac{cos(C)}{sin(C)} = 1 + \frac{sin(C)}{cos(C)}$

$\displaystyle cos^2(C) = 1 + sin^2(C)$

$\displaystyle C = n* \pi$

$\displaystyle cos^2(0) = 1 + sin^2(0)$

1 = 1

I don't know if this is proof that y(0) = 0