Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
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
add comment

2 Answers

up vote 0 down vote accepted

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

share|improve this answer
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
show 5 more comments

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.