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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.2012-02-01

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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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    You have arbitrary constants $C_1$ and $C_2$ in your solution. Putting in $x=0$, $y=0$, $z=0$ gives you two (linear) equations in those two symbols, which you can solve. Accepting the answer is good. Upvoting the answer by clicking the uparrow is also good, but you need 15 points before you can do that.2012-02-02
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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 calculat$i$ng only constants in Yp and/or Zp is enough. How to calculate the ones in Xp?2012-02-02