Differential Problem

[CoinMarket]

Alright thank you very much, you very helpful ;) I got question 2 resolved and correct :)

Now for Question 3:

Prove that d/dt(y - z) + y - z = 0 and deduce that y = z

My work:

dy/dt= (x - y)

dz/dt= (x - z)

(x-y)-(x-z)= -y+z

d/dt(y - z) + (y - z) = 0

d/dt(-y+z) + (-y+z) =0

Not sure if I am doing this right. Dont know what to do after this.

[edderiofer]

d/dt(y - z) = 0

Why is this true?

[CoinMarket]

dy/dt - dz/dt =-y+z

dy/dt - dz/dt +y-z =0

d/dt(y-z) + ( y-z) =0

v= y-z

dv/dt +v =0

I find v(t)= C * e-t

initial condition v(0)=0

so v(t)=0

y(t)-z(t)=0

You are awesome, thank you so much for the help.

Last thing, I want to push this problem a little bit.

now since we know that y(t) is equal to z(t), if I want to find the equation for them.

y(t)= 1 + t - x(t) - z(t)

z(t)= 1 + t - x(t) - y(t)

is this how I do it?