[SOLVED] A differential equation

• Jul 20th 2008, 09:13 PM
arbolis
[SOLVED] A differential equation
I must do something wrong. I have to solve $f'(t)=t^2f^2(t)$ with the initial condition $f(0)=\frac{1}{2}$.
So I wrote $\frac{df}{dt}=t^2f^2(t) \Leftrightarrow \frac{df}{f^2(t)}=t^2dt \Leftrightarrow \int \frac{df}{f^2(t)dt}=$ $\int t^2 dt \Leftrightarrow -\frac{1}{f(t)}=\frac{x^3}{3}+C \Leftrightarrow f(t)=-\frac{3}{t^3}+C$. I cleary see that $f$ in $0$ don't exist, so how could it be equal to $\frac{1}{2}$, since whatever value of $C$ won't satisfy the impossible.
• Jul 20th 2008, 09:16 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
-\frac{1}{f(t)}=\frac{x^3}{3}+C \Leftrightarrow f(t)=-\frac{3}{t^3}+C[/tex].

If $\frac{1}{f(t)} = \frac{t^3}{3} + C$ then $f(t) = \frac{3}{t^3+3C} = \frac{3}{t^3+k}$.
• Jul 20th 2008, 09:19 PM
kalagota
Quote:

Originally Posted by arbolis
I must do something wrong. I have to solve $f'(t)=t^2f^2(t)$ with the initial condition $f(0)=\frac{1}{2}$.
So I wrote $\frac{df}{dt}=t^2f^2(t) \Leftrightarrow \frac{df}{f^2(t)}=t^2dt \Leftrightarrow \int \frac{df}{f^2(t)dt}=$ $\int t^2 dt \Leftrightarrow -\frac{1}{f(t)}=\frac{x^3}{3}+C \Leftrightarrow f(t)=-\frac{3}{t^3}+C$. I cleary see that $f$ in $0$ don't exist, so how could it be equal to $\frac{1}{2}$, since whatever value of $C$ won't satisfy the impossible.

from $-\frac{1}{f(t)}=\frac{x^3}{3}+C$:

we have $f(t) = -\frac{1}{\frac{t^3}{3}+C}$

this is different from $f(t)=-\frac{3}{t^3}+C$