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

[wshfulthinker]

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

 

[Mark44]

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

If f(x) had been 32x²e^-x, you would try a particular solution of the form y_p = 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.

 

[wshfulthinker]

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.

 

[HallsofIvy]

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