# Can I isolate the function in this equation?

I need help with an equation for a project I’m working on :

I need to isolate the function $f()$ in what follows so I can put it in a software like maple or wolfram.

I have a function $p$, that monotonically decreases (the division at the end serves to normalize). $p= 0,1*e^{(-0.003x)}/p(0)$

The function that interest me, $f()$, is informally given as follows: $f(0)=0$. At each time instant t, the function grows by $1%$ of the value of $p(t)$ minus the cumulative $f(t)$ so far. For formally: $\Delta f(t)/dt = 0.01*( p(t) - \int_0^t f(t) dt ).$

where $\Delta f(t)/dt$ is the variation of function $f()$ at time $t$.

Is there any way to isolate the function $f(t)$ in this equation? Ideally, can it be stated only in terms of $p(t)$ instead of in a recursive manner?

• Take the derivative wrt t – InertialObserver Sep 3 '18 at 23:22
• @Rafa Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE – gimusi Oct 23 '18 at 20:58

We should write

$$f'(t)=\frac{df(t)}{dt} = 0.01\cdot\left( p(t) - \int_0^t f(u) du \right)$$

from which we obtain

$$\implies f''(t)= 0.01\cdot\left( p'(t) - f(t) \right) \implies f''(t)-0.01\cdot f(t)=0.01\cdot p'(t)$$

that is a linear differential equation.

Define $F(t)$ as the antiderivative of $f(t)$. Thus, the equation simplifies to $$F''(t)=\frac{1}{100}(p(t)-\int_0^tf(t)dt)=\frac{1}{100}(p(t)-F(t)+F(0)).$$

Using $p(t)=e^{\frac{-3}{1000}t}$ (I assume this is what you meant), this simplifies down to $$F''(t)=\frac{1}{100}(e^{\frac{-3}{1000}t}-F(t)+F(0)).$$

There are many techniques to solve this, but in the end we get $$F(t)=F(0)+c_1\sin(\frac{t}{10})+c_2\cos(\frac{t}{10})+\frac{10000}{10009}e^{\frac{-3}{1000}t}$$

(I found this using wolfram alpha). Of course, we want $f(t)$ and not $F(t)$. Simply differentiating gives us $$f(t)=\frac{1}{10}c_1\cos(\frac{t}{10})-\frac{1}{10}c_2\sin(\frac{t}{10})-\frac{30}{10009}e^{\frac{-3}{1000}t}.$$

Inputting $f(0)=0$, we find $$0=f(0)=\frac{1}{10}c_1-\frac{30}{10009}\Rightarrow c_1=\frac{300}{10009}.$$

Unfortunately, we still need another constant and the most simplified equation for $f(t)$ we can get is $$f(t)=\frac{30}{10009}\cos(\frac{t}{10})-C\sin(\frac{t}{10})-\frac{30}{10009}e^{\frac{-3}{1000}t}$$

for some constant $C$.

• hmmm Maybe I'm going at this the worng way. Is there a way to state what I'm looking for :" f() grows as at a rate of 1% of p() minus its own cumulative value" in a non-recursive manner that does not involve f()? – Rafa Sep 4 '18 at 0:59
• or, is there a reasonable way to approximate it? – Rafa Sep 4 '18 at 1:21