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$$ y'+3y+4z=2x $$ $$ z'-y-z=x $$ 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

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$y=z'-z-x$. $y'=z''-z'-1$. $(z''-z'-1)+3(z'-z-x)+4z=2x$. So now you have a 2nd order linear constant coefficient inhomogeneous equation for $z$. Can you solve it?

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Oh thank you. I think I can. I'll see it in a few minutes. –  aarnes Feb 1 '12 at 23:43
    
Ok, this time the solution is closer to the one on wolfram alpha but still not the same. y=(-2C1+C2-2C2x)e^{-x}-6x+14 and z=(C1+C2x)e^{-x}+5x-9 –  aarnes Feb 1 '12 at 23:56
    
Here's what I do: After your tansformation into second order linear inhomogeneous ODE I get z''+2z'+z=5x+1. From that point I get complementary solution to homogeneous part z''+2z'+z=0 z (comp.) = C1e^{-x}+C2xe^{-x}. I put x in second term because of result from homogeneous eq. Then I say that particular solution is z (part) = AX+B, z (part)' = A, z (part)'' = 0. From that I get A=5 and B=-9 . It would seem that particular solutions are correct. Then I calculate a derivative of z = z (comp) + z (part) and use one of the equation to calculate y. –  aarnes Feb 2 '12 at 0:09
    
It's rather late in my timezone. If you post something I'll see it in the morning. Thank you for your input! –  aarnes Feb 2 '12 at 0:11
    
What you have written looks good to me. –  Gerry Myerson Feb 2 '12 at 0:44

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.

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I'm sorry. I don't understand how to do that. I though calculating only constants in Yp and/or Zp is enough. How to calculate the ones in Xp? –  aarnes Feb 2 '12 at 9:39

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