1. ## Seperable Differential Equation

Solve , using separation of variables, given the inital condition .

I am not sure how to do this one. I tried to use trig substitution but it didnt work so I am not sure if that was the right path or not.
Thanks
AC

2. Originally Posted by Casas4
Solve , using separation of variables, given the inital condition .

I am not sure how to do this one. I tried to use trig substitution but it didnt work so I am not sure if that was the right path or not.
Thanks
AC
$(t^2 + 16)\,\frac{dx}{dt} = x^2 + 16$

$\left(\frac{1}{x^2 + 16}\right)\frac{dx}{dt} = \frac{1}{t^2 + 16}$

$\int{\left(\frac{1}{x^2 + 16}\right)\frac{dx}{dt}\,dt} = \int{\frac{1}{t^2 + 16}\,dt}$

$\int{\frac{1}{x^2 + 16}\,dx} = \int{\frac{1}{t^2 + 16}\,dt}$

$\frac{1}{4}\arctan{\frac{x}{4}} + C_1 = \frac{1}{4}\arctan{\frac{t}{4}} + C_2$

$\frac{1}{4}\arctan{\frac{x}{4}} = \frac{1}{4}\arctan{\frac{t}{4}} + C_2 - C_1$

$\arctan{\frac{x}{4}} = \arctan{\frac{t}{4}} + C$, where $C = 4(C_2 - C_1)$

$\frac{x}{4} = \tan{\left(\arctan{\frac{t}{4}} + C\right)}$

$x = 4\tan{\left(\arctan{\frac{t}{4}} + C\right)}$

And using the initial condition $x(0) = 4$

$4 = 4\tan{\left(\arctan{\frac{0}{4}} + C\right)}$

$4 = 4\tan{C}$

$1 = \tan{C}$

$C = \arctan{1} = \frac{\sqrt{2}}{2}$.

Therefore $x = 4\tan{\left[\arctan{\left(\frac{t}{4}\right)} + \frac{\sqrt{2}}{2}\right]}$

3. Yes , then use the trig. formula

$\tan(A + B ) = \frac{ \tan(A) + \tan(B) }{ 1- \tan(A) \tan(B) }$

$x = 4 \frac{ 1 + t/4}{1- t/4} = 4 \frac{ 4 + t}{4-t}$