right OK, the mass M at time t of the leaves of a certain plant varies according to the differential equation

dM/dt=M-M^2

a given that at time t=0, M=0.5 find an expression for M in terms of t

here is my wrong answer
(intregral) 1/M*dM=(integral) 1-M*dt

lnM=t-Mt

M=e^t * e^-mt

(the right answer is M=(e^t)/1+e^t)

how does it get to this?

Since you gave this an earnest effort, I will show you how they got that.

\frac{dM}{dt}=M-M^{2}

Separate variables:

\frac{dM}{M-M^{2}}=dt

Integrate:

\int\frac{dM}{M-M^{2}}=\int{dt}

ln(\frac{M}{M-1})=t+c

\frac{M}{M-1}=e^{t}e^{c}

e^{c} is a constant we'll call C:

\frac{M}{M-1}=Ce^{t}

Solve for M=\frac{Ce^{t}}{Ce^{t}-1}

Using the IC M=1/2 when t=0:

\frac{1}{2}=\frac{Ce^{0}}{Ce^{0}-1}

Solving for C, we find C=-1

So, we have:

M=\frac{-e^{t}}{-e^{t}-1}=\frac{e^{t}}{1+e^{t}}

Does this help? Follow that?

ln(m-1)=t+c I don't understand how you obtained the left hand side

or
M=(ce^t)/(ce^t)-1 part from the stage before it

the rest I understand thanks

I just integrated and solved for M.

\frac{1}{M-M^{2}}=\frac{1}{M}-\frac{1}{M-1}

Now, integrating, we get ln(M)-ln(M-1)

Which, by the properties of logs, equals:

ln(\frac{M}{M-1})

Then, I just did the e thing and solved for M. Just algebra.

right so you split it up to get logs OK, could you explaine the

M=(ce^t)/(ce^t)-1 part and how to get to it from the stage before it please

(also I'm running into barriers with my other question ill post my working on that page if you could tell me where I've gone wrong please)

It's just algebra, albear.

\frac{M}{M-1}=Ce^{t}

But \frac{M}{M-1}=\frac{1}{M-1}+1

\frac{1}{M-1}+1=Ce^{t}

\frac{1}{M-1}=Ce^{t}-1

M-1=\frac{1}{Ce^{t}-1}

M=\frac{1}{Ce^{t}-1}+1=\frac{Ce^{t}}{Ce^{t}-1}

Ahh OK thanks

You're welcome. I hope you learned a few littles things. Practice your algebra. It is the crux of all further math study.