System of differential equations using Laplace transform

Question

Using Laplace transform, solve the system:

$w'+y=\sin(x)$

$y'-z=e^x$

$z'+w+y=1$

where $w(0)=0$ and $z(0)=y(0)=1$.

Answer 1

Are w, y and z functions of x?

If so, hint:

$w'' + y' = \cos x$ from eq 1

$\to w'' + e^x + z = \cos x$

$\to w''' + e^x + z' = -\sin x$

$\to w''' + e^x + (1-w-y) = -\sin x$

$\to w''' - w = - \sin x - e^x - 1 + y$

$\to w''' - w = - \sin x - e^x - 1 + (\sin(x) - w')$

$\to w''' + w' - w = - \sin x - e^x - 1 + \sin(x)$

$\to w''' + w' - w =- e^x - 1$

Can you apply Laplace transform now?

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Apply Laplace to get:

$s^3 F(s) - s^2 w(0) - s' w'(0) - s^0 w''(0) + s' F(s) - s^0 w(0) - F(s) = -L(e^x) - L(1)$ (I think. Haven't done Laplace in awhile)

Get $L(e^x)$.

Get L(1).

Solve for F(s) and then:

http://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx