anonymous
  • anonymous
y''+3y'+2y=4e^x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
What have you got so far on this problem? Do you have the complimentary solution? A particular solution?
anonymous
  • anonymous
I got nothing.
anonymous
  • anonymous
Okay, so we first need the complimentary solution. This means we need to solve \[ y'' + 3y' + 2y = 0 \] Do you recall how to solve the homogeneous differential equation?

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anonymous
  • anonymous
i don't remember everything, but the x= -1 and -2. is that correct?
anonymous
  • anonymous
Yes, that is correct. Now, with those you can write down the general solution to the homogeneous differential equation. Recall that if you have two real roots to the characteristic equation, say \(r_{1}\) and \(r_{2}|) then the general solution is, \[ y_{c}(t) = c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t} \] So, for this case the complimentary solution is, \[ y_{c}(t) = c_{1}e^{-t}+c_{2}e^{-2t} \] Now, we need a particular solution. You can do either Undetermined Coefficients (probably the easiest in this case) or Variation of Parameters. Recall either of those?
anonymous
  • anonymous
Arg, messed up the math input.... The general solution should be, \[ y_{c}(t) = c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t} \] and for this case the actual soluton is, \[ y_{c}(t) = c_{1}e^{-t}+c_{2}e^{-2t} \]
anonymous
  • anonymous
-2e^-6t
anonymous
  • anonymous
Nope. Which method did you use to get that particular solution? Easier to help if I know the method you used... Also, just realized you were using x's instead of t's so all my t's above should really be x's. I'm to used to working with t's in these and I just naturally go with them....
anonymous
  • anonymous
i used wronski. But how do i use Undetermined Coefficients in this case?
anonymous
  • anonymous
Okay, with Undetermined Coefficients all we need to do is "guess" at the form of the particular solution. Because we have an exponential on the right side and we know that those don't just appear in the differentiation process it might (hopefully :) ) make sense that the form of the particular solution should be, \[ Y_{p} = A e^{x}\] We don't know what the coefficient would have to be at this point so we just call it A. Once we have the guess all we need to do is "plug" the guess into the differential equation. In otherwords, all we need to do is take a couple of derivatives of the guess, plug those and the guess itself into the appropriate places on the left and simply the left side. Once you've done that simply compare the two sides. We know that that they shouuld be the same (provided we guessed correctly of course....) so it should be fairly simple to see what value A would need to be. Do that and let me know what you get or if you get stuck let me know how far you got in the process.
anonymous
  • anonymous
i'm totally stuck on this task. can you show me the answer?
anonymous
  • anonymous
I generally don't like to just give answers. It's best if I help you arrive at the answer so you can learn the process. So, let's go through this step by step. I'm assumnig that you agree that \(Y_{p}^{\prime} = Ae^{x}\) and \(Y_{p}^{\prime \prime} = Ae^{x}\). Correct? Now, "plug" \(Y_{p}\) into the differential equation. I.e., \[ Y_{p}^{\prime \prime} + 3 Y_{p}^{\prime} + 2Y_{p} = 4e^{x} \] If you plug in all the formulas for the derivatives and the guess into the left side what do you get?
anonymous
  • anonymous
Ae^x+3Ae^x+2Ae^x=4e^x ???
anonymous
  • anonymous
Yep! Now, because all the terms on the left are coefficients times \(e^{x}\) we can combine them into a single term right? What do you get if you do that?
anonymous
  • anonymous
6Ae^x=4e^x
anonymous
  • anonymous
Perfect! Now, if we guessed correctly, the two sides should be the same. In otherwords, since there is only an \(e^{x}\) on both sides their coefficients should be the same. So, all we need to do is set the coefficients equal and this will give us an equation that we can solve for A. When you do that what do you get?
anonymous
  • anonymous
A=\[2\div3\] y(x)=\[2\div3\]e^x
anonymous
  • anonymous
Correct! So, we now know the complimentary and particular solution of the differential equation. \[ y_{c}(x) = c_{1}e^{-x} + c_{2}e^{-2x}\] \[Y_{p}(x) = \frac{2}{3}e^{x}\] Do you recall how to combine those to get the solution to the differential equation?
anonymous
  • anonymous
no
anonymous
  • anonymous
Okay, the general solution to the differential equation is then just \[y(x) = y_{c}(x) + Y_{p}(x)\] So, add the two together and you're done!
anonymous
  • anonymous
Thank you so much for your help:)
anonymous
  • anonymous
No problem!

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