# Help with a system of linear first-order ODEs using elimination method

$$y'+3y+4z=2x$$ $$z'-y-z=x$$ x is independent variable!

The solution I get is not the same as the one on Wolfram Alpha http://www.wolframalpha.com/input/?i=y%27%2B3y%2B4z%3D2x%2C+z%27-y-z%3Dx . So how to solve it? My solutions are: $$y=C1e^{-x}+C2xe^{-x}-6x+10$$

$$z=-(C1/2)e^{-x}-(C2/4)e^{-x}-(C2/2)xe^{-x}+2x$$

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 Did you try checking your work by substituting your solutions in to the differential equations? Your homogeneous terms (the ones containing $C1$ and $C2$) are correct, but the particular solution $y = -6x+10$, $z=2x$ doesn't satisfy either equation. – Robert Israel Feb 1 '12 at 23:40

$y=z'-z-x$. $y'=z''-z'-1$. $(z''-z'-1)+3(z'-z-x)+4z=2x$. So now you have a 2nd order linear constant coefficient inhomogeneous equation for $z$. Can you solve it?
For a particular solution, try substituting $y = a x + b$, $z = c x + d$ in to the differential equations, and find the constants $a,b,c,d$ that make the resulting equations true. Note that the coefficients of $x$ involve only $a$ and $c$, so first solve for those.