Finding Particular Solutions of y''+10y'+25y= 32xe^(-x)

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Homework Help Overview

The discussion revolves around finding a particular solution to the differential equation y'' + 10y' + 25y = 32xe^(-x). The original poster has already determined the general solution for the homogeneous case and is now focused on the non-homogeneous part.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the form of the particular solution, suggesting variations based on the nature of the non-homogeneous term. The original poster attempts a specific form but questions its validity after receiving feedback.

Discussion Status

Some participants have provided guidance on the appropriate form for the particular solution, indicating that adjustments may be necessary based on the structure of the non-homogeneous term. The original poster acknowledges a successful attempt based on this feedback, but there is no explicit consensus on the best approach.

Contextual Notes

There is a note regarding the appropriateness of the forum section for this topic, suggesting that the original poster may have misclassified their question. Additionally, the discussion touches on the general approach to choosing particular solutions based on the degree of polynomial terms in the non-homogeneous part.

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Homework Statement


Consider the differential equation:
y''+10y'+25y= f(x)

Find a particular solution if f(x) = 32xe^(-x)

Homework Equations


I already did the general solution when f(x)=0 and that is Ae^(-5x) + Bxe^(-5x)


The Attempt at a Solution


I tried yp=axe^(-x) and got a= 4x+2 which is wrong

The answer is (2x-1)e^(-x)

does anyone know what particular solution i can try in order to get the answer?

Thanks
 
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If f(x) had been 32e-x, you would want to try yp = Ae-x. Since f(x) = 32xe-x, you want your particular solution to be yp = Ae-x + Bxe-x.

If f(x) had been 32x2e-x, you would try a particular solution of the form yp = Ae-x + Bxe-x + Cx2e-x. There's a reason for all of this, but I'll leave that for your instructor.

BTW, this is hardly a Precalculus question. You should have posted it in Calculus and Beyond.
 


Hi, thankyou so much for your reply. I tried it and it worked! i shall write it down and remember that forever now!

Also, sorry about posting in the wrong section! I can't believe i did that because i took so long to check that my post was right.. i guess i forgot to check if i had clicked on the right section..! Thankyou so much though.
 


Generally speaking when a "right hand side" involves an [itex]n^{th}[/itex] power of x, you should try a polynomial of degree n down.
 

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