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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?

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  • $\begingroup$ 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. $\endgroup$ – abel Jan 8 '15 at 20:42
  • $\begingroup$ the solution is $y(x)=3e^{2x}$ $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '15 at 20:44
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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?

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

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