Differential Equations, Particular and Complimentary solutions

[JefeNorte]

Homework Statement

Given the differential equation for y=y(x)
(1) L[y]=y"+2by'+yb^2=(e^(-bx))/(x^2) x>0
a)find the complementary solution of (1) by solving L[y]=0
b)Solve (1) by introducing the transformation y(x)=(e^(-bx))*v(x) into (1) and obtaining and solving completely a differential equation for v(x). Use this to identify the particular solution

The Attempt at a Solution

Follwing the steps outline at http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
I converted y"+by'+yb^2 to r^2+2br+b^2=0

By factoring I determined that r1=r2=-b
so the complimentary solution should be y=c1*e^(-bt)+c2*e^(-bt)

Is this the right way to solve for a complimentary solution? If so how do I "induce the transformation to solve fro the particular solution?"

 

[HallsofIvy]

No, that is not correct. the complimentary solution is y= C_1e^{-bt}+ C_2t e^{-bt}.

How about doing what the problem asks you to do: "Solve (1) by introducing the transformation y(x)=(e^(-bx))*v(x) into (1)"?

If y(x)= e^{-bx}v(x), what are y' and y"? Put those into the equation and see what equation you get for v.

Don't forget that the equation is 2by', not the by' you wrote in "I converted---". With the "2", it reduces nicely. Without, it's a mess!

 

[JefeNorte]

So after introducing y(x)= v(x)*e^(-bx) to the initial equation I came up with v"=1/(x^2)

Using this I came up with v(x)=(4/3)x^(1.5)

So y(x)=((4/3)x^1.5)*e^(-bx) is the particular solution? I apologize for being so dense but I don't actually take this class until next semester, I am just trying to get ahead. Thanks