0
$\begingroup$

$$ y'+3y+4z=2x $$ $$ z'-y-z=x $$ 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: $$ y=C1e^{-x}+C2xe^{-x}-6x+10 $$

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

  • 0
    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

2 Answers 2

0

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

  • 0
    Oh thank you. I think I can. I'll see it in a few minutes.2012-02-01
  • 0
    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-92012-02-01
  • 0
    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.2012-02-02
  • 0
    It's rather late in my timezone. If you post something I'll see it in the morning. Thank you for your input!2012-02-02
  • 0
    What you have written looks good to me.2012-02-02
  • 0
    So why isn't my solution same as the one on Wolfram Alpha? Could it be their algorithms are wrong?2012-02-02
  • 0
    Have you tried checking both solutions, yours and Wolfram's, to see whether they work? That's the great thing about differential equations, you know: you can always tell whether an answer is correct, simply by plugging it back into the equation(s) to see if it works.2012-02-02
  • 0
    Gerry, I checked my solutions and they're correct. Why is wolfram then showing those solutions I'll never know. Maybe an algorithm problem. The good news is that solutions when put inside main equations "fit". 2X=2X and x=x for the second. So thank you! Do I just click accept answer or is there a way to reward you through some mechanism? I clicked the "correct" sign.2012-02-02
  • 0
    Would it be too much to ask what to do if one has initial condition for this system? For example: x=0, y=0, z=0.2012-02-02
  • 0
    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
0

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.

  • 0
    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?2012-02-02