Integration (help please)

[edumat]

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) ]

 

[DrPhil]

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) ]

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.

 

[edumat]

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.

 

[mathmari]

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)...

 

[Deleted member 4993]

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.

 

[edumat]

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?

$\displaystyle f'(x) = -\frac{f(x)}{\sqrt{16-f(x)²}}$; $\displaystyle f(0) = 4$

 

[Deleted member 4993]

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.