How to solve a system of linear differential equations using laplace transformation when one of the initial condition is not at zero?

Question

A new user here!

For specifics, the problem is: $$x' - y'' = - 2 \cosh t$$ $$y' - x'' = 2 \sinh t$$

$$y(0) = y'(0) = y''(0) = 1$$ $$x(1) = - x'(1) =x'' (1) = 1/e$$

From what I know, I need to change $x(1)$ to $x(0)$, by letting a variable $u = t-1$. But if I do that, $y(0)$ would change to $y(-1)$.

I am at a loss to what to do. No need to solve the whole problem, just point me to the right direction. Thanks!

Answer 1

Differentiate the first equation and add the second equation to it to obtain

$$x''-y'''+(y'-x'')=-2\sinh t+2\sinh t.$$

$$\implies y'''-y'=0$$

Now, use the Laplace transform. After having obtained the solution for $y(t)$ you can plug this into the second ode and then do the substitution in time that you proposed.