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Diff Eq problem

  • Thread starter hils0005
  • Start date
  • #1
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[SOLVED] Diff Eq problem

Homework Statement


y'' - 3y' - 4y= 5e^-x - 3x^2 + 7


Homework Equations



I think I would need to find complimentary solution, then the particular solution using variation of parameters
y=y(c) + y(p)



The Attempt at a Solution



y(c)=
r^2-3r-4=0
(r-4)(r+1)=0, r=4,-1
y(c)=c(1)e^4x+c(2)e^-x

this is where I get stuck as I do not know what to use for y(p), would it be
y(p)=Axe^-x + Bx^2 + Dx +E ???
y'(p)=Ae^-x - Axe^-x + 2Bx + D
y"(p)=-Ae^-x -Ae^-x +Axe^-x +2B= Axe^-x - 2Ae^-x + 2B

(Axe^-x - 2Ae^-x + 2B) - 3(-Axe^-x + Ae^-x + 2Bx + D) -4(Axe^-x +Bx^2 + Dx +E)= (5e^-x - 3x^2 + 7)

am I even heading in the right direction?
 
Last edited:

Answers and Replies

  • #2
200
0
everything looks right so far. now you just want to collect terms and match coefficients.
 
  • #3
rock.freak667
Homework Helper
6,230
31
If I remember correctly that is not variation of parameters but method of undermined coefficients. Variation of parameters is something different.
 

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