Differential Equations Question?

Question

I am asked to express the following systems of first-order ordinary Diff Eq's in the form y ' =A(t)y +b(t) for a suitable A and b.

1) x ' = x

y' = x + y

z' = x + y + z

2) x' = tx + 1

y' = x + z

z' = 1 - tz

3) x' = tx + cos (t)

y' = cos (t)y + t

If you could help me out on any or all of the three of these that would be great. I have no clue where to even start on these problems.

Answer 1

1)

x' - x = 0

x'e^(-t) -xe^-t = 0

(-xe^-t)' = 0

-xe^(-t) = c1

-x = c1*e^t

x = -c1e^t

y' = -c1e^t + y

y' - y = -c1e^t

y'e^(-t) - ye^(-t) = -c1e^(-t)

(-ye^(-t))' = -c1e^(-t)

-ye^(-t) = c1e^(-t) + c2

-y = c1 + c2e^t

y = -c1 + c2e^t

z' = -c1e^t - c1 + c2e^t + z

z' = (c2-c1)e^t + z - c1

z' - z = (c2-c1)e^t - c1

z'e^(-t) -ze^(-t) = (c2-c1) - c1e^(-t)

(-ze^(-t))' = (c2-c1) - c1e^(-t)

-ze^(-t) = (c2-c1)t + c1e^(-t) + c3

-z = (c2-c1)te^t + c3e^t + c1

z = (c1-c2)te^t - c3e^t - c1

So all this work was a waste... from above we have

y' - y = -c1e^t

y' = y - c1e^t

Not sure what z was for, but it's solved just in case.

2)

z' = 1 - tz

z' + tz = 1

z'e^(.5t^2) + tze^(.5t^2) = e^(.5t^2)

(ze^(.5t^2))' = e^(.5t^2)

ze^(.5t^2) = erfi(x√2)√(pi)/√(2) + c1

z = erfi(x√2)√(pi)/e^(.5t^2)√(2)    + c1

Solution to be finished tomorrow. If you have no clue what the error function and imaginary error functions are, I probably went massively off somewhere.