Shifting initial conditions to 0, Differential Equations

use the method of laplace transforms to solve the given initial value problem.

x'-2x+y'=-(cost + 4sint); x(π) = 0

2x +y' + y=sint + 3cost; y(π) = 3

I'm not sure how to shift the initial conditions to 0.

You can use a change of variable u = t - π.
Then, dx/dt = dx/du * du/dt = dx/du. Similarly, dy/dt = dy/du.

Since cos t = cos(u + π) = -cos u, and sin t = sin(u + π) = -sin u,
the system of DE's transforms to

x' - 2x + y' = -(-cos u + 4 * -sin u); x(0) = 0
2x + y' + y = -sin u + 3 * -cos u; y(0) = 3

Rewrite this as
x' - 2x + y' = cos u + 4 sin u; x(0) = 0
2x + y' + y = -sin u - 3 cos u; y(0) = 3

Can you take it from here?

Let τ = t - π
τ + π = t

dx/dτ = (dx/dt)/(dτ/dt) = dx/dt and similarly for y, so the left side of the equations formally stay the same.

cos(t) = cos(τ + π) = cos(τ) cos(π) - sin(τ) sin(π) = -cos(τ)
sin(t) = sin(τ + π) = sin(τ) cos(π) + cos(&tau:) sin(π) = -sin(tau;)

Once you solve for the Laplace transforms of x and y and invert them, you will have x(τ) and y(τ); obviously, you will need to replace the τ's by (t - π). With "normal" Laplace transform problems (i.e., with initial conditions at t = 0) we tacitly assume that x(t) and y(t) are zero for t < 0. Are you told to do that here, but for t < π? If so, you'll need either to specify t > π or to multiply your solutions by U(t - π) where U(*) is the unit step (Heaviside) function.