Finding the differential equation, given a solution?

re write equation as y = (a+bx)e^-x
y/e^-x = a+bx
(y/e^-x)-bx=a
differentiate
(e^-x y' + ye^-x)/e^-2x -b =0
then (e^-x y' + ye^-x)/e^-2x = b
now differentiate again
simplifying first
e^-x (y' +y)/e^-2x =e^x (y'+y)=b
e^x (y'' + y) + (y'+y) e^x =0
e^x( y'' +2y +y)=0
since e^x never =0 we are left with
y''+2y'+y=0

The answer to your question is simple, but depends upon prior knowledge.
You might decide to record the summary of solution types included below.

Solutions of the type y = e^rx(A + Bx) are type (2) in that summary, and are
found from DEs that have auxiliary equations with equal real roots.
y = e^-x(A + Bx) derives from both real roots being r = -1
The auxiliary equation was (r + 1)^2 = r^2 + 2r + 1 = 0
The related differential equation was: y'' + 2y' + y = 0

Solutions of the type y = Ae^px + Be^qx are type (1) in that summary, and are
found from DEs that have auxiliary equations with different real roots
y= Ae^(-x) + Be^(-3x) derives from roots r = -1 and r = -3
The auxiliary equation was (r + 1)(r + 3)= r^2 + 4r + 3 = 0
The related differential equation was: y''+ 4y' + 3y = 0


SUMMARY: Solutions to the differential equation ay'' + by' + cy = 0
depend on the form of the roots of the auxiliary equation
am^2 + bm + c = 0
1) Real different m = p, q
y = Ae^px + Be^qx
2) Real equal m = r (twice)
y = e^rx(A + Bx)
3) Complex m = a +/- ib
y = e^ax[Acos (bx) + Bsin (bx)]
4) Real and same modulus but opposite sign
y'' - n^2y = 0
m^2 - n^2 = 0
m = +/- n (real)
y = Ae^nx + Be^-nx OR more usually
y = Ccosh(nx) + Dsinh(nx)
5) Imaginary roots derive from
y'' + n^2y = 0
m^2 + n^2 = 0
m = +/- ni
y = Pe^inx + Qe^-inx OR more usually
y = Acos (nx) + Bsin (nx)

Regards – Ian H

y=ae^(-x)+bxe^(-x) gives ye^x=a+bx so, differentiate with respect to x gives
y'e^x+ye^x=b and differentiate again givves
y"e^x+y'e^x+y'e^x+ye^x=0 and I leave you to finish.

Please read:
http://www.math24.net/second-order-linear-homogeneous-differential-equations-with-constant-coefficients.html

(1)
y
= a*e^(-x) + bx*e^(-x)
= ( bx + a ) * e^(-x)

So, the roots of the characteristic eq. are - 1 , - 1
( k + 1 )( k + 1 ) = 0
k^2 + 2k + 1 = 0

Thus, the differential eq. is y'' + 2y' + y = 0

(2)
y = a*e^(-x) + b*e^(-3x)

So, the roots of the characteristic eq. are - 1 , - 3
( k + 1 )( k + 3 ) = 0
k^2 + 4k + 3 = 0

Thus, the differential eq. is y'' + 4y' + 3y = 0