# Thread: First Order Ordinary differential Equation

1. ## First Order Ordinary differential Equation

I have the first order Initial Value Problem ODE of

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

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

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

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

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

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

y(0) = 0 so

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

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

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

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

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

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

$C = n* \pi$

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

1 = 1

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

2. Originally Posted by Fallen186
I have the first order Initial Value Problem ODE of

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

I calculate that $y' = sec^2(2x+c)$
You have this as your LHS.

now evaluate $1 + 4y^2$ to get the same.

$y(0)=0$ is used to solve for C.

3. Life is complex: it has both real and imaginary components.
Try graphing it

4. Originally Posted by Fallen186
I have the first order Initial Value Problem ODE of

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

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

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

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

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

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

y(0) = 0 so

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

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

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

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

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

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

$C = n* \pi$

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

1 = 1

I don't know if this is proof that y(0) = 0
$\frac{dy}{dx} = 1 + 4y^2$

$\frac{dx}{dy} = \frac{1}{1 + 4y^2}$

$x = \int{\frac{1}{1 + 4y^2}\,dy}$

$x = \frac{1}{2}\arctan{(2y)} + C$

$x - C = \frac{1}{2}\arctan{(2y)}$

$2(x - C) = \arctan{(2y)}$

$2y = \tan{2(x - C)}$

$y = \frac{1}{2}\tan{2(x - C)}$.

You are told that $y(0) = 0$, so

$0 = \frac{1}{2}\tan{2(0 - C)}$

$0 = \frac{1}{2}\tan{(-C)}$

$\tan{(-C)} = 0$

$C = 0$.

Thus $y = \frac{1}{2}\tan{(2x)}$.