MHB How Do You Rearrange dS/dt = (n-x-rS)(L-1) to Solve for t?

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To rearrange the equation dS/dt = (n-x-rS)(L-1) to solve for t, separate the variables and integrate. The integration involves the left side being ∫(dS/(n-x-rS)) and the right side as ∫(L-1) dt. This process allows for the determination of t in terms of S and other variables. The integration will yield an expression that can be manipulated to isolate t. This method provides a systematic approach to solving the differential equation.
BSLAHi
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Hello,

I have been given a general equation that is dS/dt = (n-x-rS)(L-1)
and it needs to be rearranged in order to for the subject to be t.

I have spent a while trying to find a way to attempt this but with no luck
so I will leave this here.

Thanking You,
BSLAHi
 
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Hello BSLAHi.

You can separate the variables and integrate.
$$\frac{\mathrm dS}{\mathrm dt}\ =\ (n-x-rS)(L-1)$$
$\displaystyle\implies\ \int\frac{\mathrm dS}{n-x-rS}\ =\ \int(L-1)\,\mathrm dt$.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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