# Ordinary differential equation with Boundary value conditions

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?

• derivative of $[f(t)]^2$ is $2f(t)f^\prime(t)$ not $2f(t)$ as you seem to have assumed. look up separation of variables or exponential growth models. – abel Jan 8 '15 at 20:42
• the solution is $y(x)=3e^{2x}$ – Dr. Sonnhard Graubner Jan 8 '15 at 20:44

You have a separable differential equation $$\frac{f'(t)}{f(t)}=2.$$ Integrating from $t=0$ to $t=1$, you get $$\ln f(1)-\ln f(0)=\int_0^1\frac{f'(t)}{f(t)}\,dt = \int_0^1 2\,dt=2.$$ I suppose you can take it from here?

Let's assume $f(t) = c_1e^{c_2t}$

It's given that $f(0) = 3$ so $c_1 = 3$.

Differentiating gives $f'(t) = c_1c_2e^{c_2t}$

Plugging in the differential equation: $f′(t)=2f(t)$ gives $c_1c_2e^{c_2t} = 2c_1e^{c_2t}$

Cancelling the $c_1$ and the $e^{c_2t}$ gives $c_2 = 2$.

Thus: $f(t) = 3e^{2t}$