Hello,

I'm pretty stumped on this problem at the moment.
What I initially tried was zero-state and zero-input but I'm not confident that is the correct method.


What I devised for x(t) is x(t) = u(t) -(1+e^(-t))u(t-1)
*where u(t) is the unitstep function.

Is it possible for me to solve the differential equation with seperate inputs?
So, suppose I solve it for y'+2y = 1 (0 < t < 1)
and y' +2y = e^(-t) (t>1)

 

So, suppose I solve it for y'+2y = 1 (0 < t < 1)
and y' +2y = e^(-t) (t>1)

y'+2y = 1 (0 < t < 1)
y' +2y = e^-(t-1) (t>1)
I think for t = 1 to be x(t)=-1

y'+2y = 1
https://www.wolframalpha.com/input/?i=y'+++2y+=+1
y(t)=c*e^(-2t)+1/2


y' +2y = e^-(t-1)
https://www.wolframalpha.com/input/?i=y'+++2y+=+e^-(x-1)
y(t)=c*e^(-2t)+e^(1-t)

the simplest case:
y' +2y = 0

Exercise is not clear for me.
y' is the derivative of the function y with x or t?
y'=y'(x)=dy/dx or y'=y'(t)=dy/dt?

 

I'm under the impression that y=y(t).

Thanks for the help, trying different ways to check my answer, not allowed to use laplace.

actually, the lower peak at 1 is -1/e

 

Hi,

Then your second function part is (-1/e)*e^(-t).

When you solve the second part you have to use the final conditions from the first part as the initial conditions of the second part.
For example, say after the first part you have y=0.433, then you have to use y=0.433 as the initial conditions when you go to solve the second part. For the first part the initial condition is y=0 as they state.

Note this can also be viewed as dy/dt=x(t)-2*y in case that helps, where x(t) is viewed as the input and -2*y is viewed as the feedback.
We can then think of the solution as being the sum of all delta t times dy/dt, where delta t is some small time value.

 

Certainly, you know how to solve such an equation:
y'+2y = 1

we consider:

The result (=1) is a constant (independent of t)
y may be this constant.
if y' will be constant then y would depend on t and the result would be not a constant.
So, y=ct => y'=0 =>
0+2y=1 => y=1/2 => y2=1/2
but y=y1+y2=...........

 

Hi,

I dont think the general solution to:
y'+2*y=x(t)

is that difficult, so the general solution can be used for both parts after substituting in either x(t)=1 or x(t)=(-1/e)*e^(-t) and applying appropriate initial conditions.

If he wants to he can solve them separately though:
y'+2*y=1, with y(0)=0 for t=0 to t=1, and solution y(1) at t=1,

and:
y'+2*y=(-1/e)*e^(-t), with y(0)=y(1) from above and for t=0 to infinity and solution y(t) or y(t-1) for all time t>=1.

This is electrically equivalent to an RC low pass filter with all inputs halved and RC time constant equal to 1/2,
and inputs as noted for x(t) in this problem (except halved).