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I'm attempting to solve for $y$ as a function of $x$ if $y''''-12y'''+36y''=256e^{-2x}$, with initial conditions: $y(0)=5, y'(0)=-1, y''(0)=40, y'''(0)=-8$. I've tried to solve this multiple times and each time the solution I obtain is the same, although it's apparently incorrect. Here's my work so far:

$y(x)=y_h(x)+y_p(x)$. So first we solve for the homogeneous solution of $y''''-12y'''+36y''=0$. Let $y=e^{mx}$, and then we have $m^4-12m^3+36m^2=0 \implies m=0,0,6,6$. So our solution is of the form $y_h(x)=c_1e^{0x}+c_2xe^{0x}+c_3e^{6x}+c_4xe^{6x}$. Now we use initial conditions to solve for the constants:

$y(0)=5 \implies c_1+c_3=5$

$y_h'=c_2+6c_3e^{6x}+c_4e^{6x}+6c_4xe^{6x} \implies y'(0)=c_2+6c_3+c_4=-1$

$y_h''=36c_3e^{6x}+6c_4e^{6x}+6c_4e^{6x}+36c_4xe^{6x} \implies y''(0)=36c_3+12c_4=40$

$y_h'''=216c_3e^{6x}+36c_4e^{6x}+36c_4e^{6x}+36c_4e^{6x}=216c_4xe^{6x}\implies y'''(0)=216c_3+108c_4=-8$

Now we have a system of four equations, four unknowns, and solving , we find that $c_1=\frac{43}{27}, c_2=\frac{-131}{9}, c_3=\frac{92}{27}, c_4=\frac{-62}{9}$ (I have verified this with WolframAlpha) and so $y_h(x)=\frac{43}{27}-\frac{131}{9}x+\frac{92}{27}e^{6x}-\frac{62}{9}xe^{6x}$

Now to solve for the particular solution $y_p(x)$ we use the method of undetermined coefficients: $y_p$ is of the form $ae^{-2x}$. $y_p''=4ae^{-2x}, y_p'''=-8ae^{-2x}, y_p''''=16ae^{-2x}$. Plugging these into the D.E.:

$16ae^{-2x}-12\cdot-8ae^{-2x}+36\cdot4ae^{-2x}=256ae^{-2x}=256e^{-2x} \implies a=1$

Thus $y(x)=\frac{43}{27}-\frac{131}{9}x+\frac{92}{27}e^{6x}-\frac{62}{9}xe^{6x} + e^{-2x}$

However, according to the program we use for assignments, this incorrect. If anyone could find my error it would be greatly appreciated.

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    $\begingroup$ Your (first) mistake is in the form of $y_h$. Since $0$ is a repeated root, you need an $xe^{0x}$ term. But a more important mistake is that you shouldn't solve for the coefficients of $y_h$ until you have combined it with $y_p$, as it is the sum that must satisfy those initial conditions. $\endgroup$ – Gerry Myerson Mar 1 '15 at 8:06
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    $\begingroup$ @GerryMyerson Forgot that in the beginning, but I did use that in my solution, if you look at the final solution. I had a realization: is my error in finding the coefficients only for $y_h$ and not after I'd found $y_p$? $\endgroup$ – jofl Mar 1 '15 at 8:09
  • $\begingroup$ This is the reason. If you use your last results, the conditions are not satisfied. $\endgroup$ – Claude Leibovici Mar 1 '15 at 8:20
  • $\begingroup$ Ah, silly mistake, but great to learn. Thank you. $\endgroup$ – jofl Mar 1 '15 at 8:21
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Answer: $$y=-6 x\, {{ e}^{6 x}}+3 {{ e}^{6 x}}+{{ e}^{-2 x}}-11 x+1$$

Solved with CAS Maxima.

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  • $\begingroup$ I promise you, I will download and try it :-) But please admit, it is really not a very HQ answer. I hate to click "Recommend Deletion", particularly for posts with a clearly visible goodwill. I suggest this: 1) write a lot HQ answers 2) meanwhile, if you find such, work-intensive questions, solve them with CAS Maxima and put the result in a comment. | You can also put its link into your profile, if you are active, you will have a lot of visitors. $\endgroup$ – peterh Jun 4 '18 at 21:40
  • $\begingroup$ @peterh load(odes); eq:'diff(y,x,4)-12*'diff(y,x,3)+36*'diff(y,x,2)=256*exp(-2*x); expand(odeL_ic(eq,y,x,[0,5,-1,40,-8])); $\endgroup$ – Aleksas Domarkas Jun 4 '18 at 21:59
  • $\begingroup$ The problem with it is that it is understable only for the people knowing that CAS Maxima. Also I don't know it (yet). $\endgroup$ – peterh Jun 4 '18 at 22:02
  • $\begingroup$ Maxima CAS is free computer algebra system. maxima.sourceforge.net $\endgroup$ – Aleksas Domarkas Jun 4 '18 at 22:04
  • $\begingroup$ I know, and probably all the reviewers of the site already know it :-) $\endgroup$ – peterh Jun 4 '18 at 22:06

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