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

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SUMMARY

The discussion focuses on finding a particular solution to the differential equation y'' + 10y' + 25y = 32xe^(-x). The correct particular solution is identified as (2x - 1)e^(-x). The user initially attempted yp = axe^(-x) but received incorrect results. The forum emphasizes the importance of selecting the appropriate form for yp based on the nature of f(x), suggesting that for polynomial terms multiplied by exponential functions, the particular solution should include terms corresponding to the degree of the polynomial.

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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 n^{th} power of x, you should try a polynomial of degree n down.
 

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