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Integration (help please)
- Thread starter edumat
- Start date
All you have to do to prove the given function is to show that f(0) = C [<-- E DITED], and that the derivative is correct. Show us what you have tried.Hello,
I'd like to have some help with this problem:
If f'(x) = A*f(x)*[ B-f(x) ], A and B constants, and f(0) = C, prove that:
f(x) = B*C / [ C + (B-C)*e^(-A*B*x) ]
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Please note that the exercise gives
\(\displaystyle f'(x) = Af(x)[B-f(x)]\) and not simply \(\displaystyle f(x)\)
So, I don't know how to integrate this function, it looks like a differencial equation. I have not reached this topic in my course yet, but my teacher handed this exercise anyway, so I think there's a simplier way to solve. This exercise is at integration techniques topic.
\(\displaystyle f'(x) = Af(x)[B-f(x)]\) and not simply \(\displaystyle f(x)\)
So, I don't know how to integrate this function, it looks like a differencial equation. I have not reached this topic in my course yet, but my teacher handed this exercise anyway, so I think there's a simplier way to solve. This exercise is at integration techniques topic.
Hello,
I'd like to have some help with this problem:
If f'(x) = A*f(x)*[ B-f(x) ], A and B constants, and f(0) = C, prove that:
f(x) = B*C / [ C + (B-C)*e^(-A*B*x) ]
\(\displaystyle f'(x)= Af(x)[ B-f(x) ] => \frac{f'(x)}{f(x)[B-f(x)]}=A => \frac{f'(x)}{B}(\frac{1}{f(x)}-\frac{1}{f(x)-B})=A => \frac{f'(x)}{Bf(x)}-\frac{f'(x)}{B(f(x)-B)}=A \)
By integrating, we have:
\(\displaystyle \frac{log(f(x))}{B}-\frac{log(f(x)-B)}{B}=Ax +c=> log(\frac{f(x)}{f(x)-B})=B(Ax +c)=> \frac{f(x)}{f(x)-B}=e^{B(Ax+c)} \)
Continue by solving for f(x)...
Last edited:
D
Deleted member 4993
Guest
Hello,
I'd like to have some help with this problem:
If f'(x) = A*f(x)*[ B-f(x) ], A and B constants, and f(0) = C, prove that:
f(x) = B*C / [ C + (B-C)*e^(-A*B*x) ]
The easiest way to solve this problem is to
Show that f(x) = B*C / [ C + (B-C)*e^(-A*B*x) ] satisfies the following conditions:
f'(x) = A*f(x)*[ B-f(x) ], A and B constants, and f(0) = C.
Because the problem statement simply says "prove that" - the above interpretation (provided by Dr. Phil] is valid.
D
Deleted member 4993
Guest
Thanks to all of you guys! Now I see that both suggested paths works...
By the way, could you just help me to start this integration?
\(\displaystyle f'(x) = -\frac{f(x)}{\sqrt{16-f(x)²}}\); \(\displaystyle f(0) = 4\)
If it is a new problem - please start a new thread.