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Help solving nonhomogeneous de

  • Thread starter xtipd
  • Start date
  • #1
7
0
Hey guys just asking for a bit of help to get me on the right track.

I have the non homogeneous de 4y" + 4y' + y = 3*x*e^x, which also has some inital conditions y(0)=0 and y'(0)=0. but i only need help with getting the particular solution.

Tried method of constant coefficients and it didnt want to work for me.
I end up using: y(p) = A*t*e^t which doent seem right when i substitute in y'(p) and y"(p) into the original equaltion.

Tried method of variation of parameters and i dont think that worked either as in the end i got y(p) = 0.

So far i get the auxilary eq: r^2 + r + 1/4 = 0
=>(r + 1/2) = 0
thus r1 = r2 = -1/2
so y(h) = c1*e^-1/2*t + c2*t*e^-1/2*t

i just need help as to what method i would use to get a the particular solution y(p). from there its gravy :)

thanks in advance
 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
With respect to the method of constant coefficients: usually whenever there's a polynomial involved, you should include terms for all powers up to the order of the polynomial. In your case, you have y = (constant)*(polynomial)*(exponential), but you only included the linear term in your polynomial - you left out the constant term. Go back and try y(x) = A(x+B)e^x.
 
  • #3
7
0
With respect to the method of constant coefficients: usually whenever there's a polynomial involved, you should include terms for all powers up to the order of the polynomial. In your case, you have y = (constant)*(polynomial)*(exponential), but you only included the linear term in your polynomial - you left out the constant term. Go back and try y(x) = A(x+B)e^x.
So obvious now you have pointed that out.

Im all over it like a bad rash now

Thanks heaps
 
  • #4
diazona
Homework Helper
2,175
6
:smile: I've made that mistake many times myself... ODEs can be annoyingly tricky.
 

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