Let $f(t)$ be the solution to the ordinary differential equation $$ f′(t)=2f(t), $$ with the boundary condition $f(0)=3$, then $f(1)=?$
Give me a hint if you do not want to provide full answer. What I did was
I have integrated above ode like so:- $$ F(t) = [f(t)]^2 + C = f(t) $$ applying boundary condition, $C$ comes out to be: $9+C=3$ giving $C=-6$
So general solution would be: $$ [f(t)]^2 - f(t) - 6 = 0 $$ How can I guess value for 2nd condition of BVP?, also there is only one constant. How is it a BVP?