# Transform a continuous time differential equation of a Low Pass Filter to discrete time

Assuming I have a linear differential equation of first order low pass filter: $$RC*y(t)'+y(t) = x(t)$$ And Assuming i know now that the input x(t) is being sampled at a rate of $T_s$

how can now switch the equation to describe in the discrete time form? I know that the answer is: $$x[n] = y[n] + RC/T_s* (y[n]-y[n-1])$$

but i can't figure how to get there. I know that there is a strong relation between DTFT and fourier transform but i couldn't figure how to use it. Thanks

The answer comes from a discrete approximation of the derivative operator.
Sometime it is called "Finite Differences".

Look at $\frac{ y \left[ n {T}_{s} \right] - y \left[ n {T}_{s} - {T}_{s} \right] }{ {T}_{s} }$, what happens when ${T}_{s} \rightarrow \infty$?
We got the definition of the drivative.

Now, since we sample at interval of ${T}_{s}$ we can agree that $y \left[ n {T}_{s} \right] = y \left[ n \right]$ and $y \left[ n {T}_{s} - {T}_{s} \right] = y \left[ n - 1 \right]$.

Do that and you get:

$$R C {y}' \left ( t \right ) + y \left ( y \right ) = x \left ( t \right ) \approx R C \frac{ y \left [ n \right ] - y \left [ n - 1 \right ] }{ {T}_{s} } + y \left [ n \right ] = x \left [ n \right ]$$

Now simple adjustment and you have what you need.