# How to get the form of a particular solution of y''-2y'+y=xe^x +4 I found the general solution to be y=c1e^x+c2xe^x , so I need to find the particular solution. If I can get the particular...

How to get the form of a particular solution of y''-2y'+y=xe^x +4

I found the general solution to be y=c1e^x+c2xe^x ,

so I need to find the particular solution. If I can get the particular solution,

I think I can fond the solution of this differentical equation.

*print*Print*list*Cite

### 1 Answer

You need to form and solve the characteristic equation for `y''-2y'+y = 0` , hence you need to solve `r^2 -2r + 1 = 0` to find the homogenous solution.

Notice that the quadratic `r^2 -2r + 1` occurs if you raise r - 1 to square such that:

`(r-1)^2 = 0 =gt r_1 = r_2 = 1`

You need to find initial trial solution hence `y = d_1x*e^x + 4d_2`

You need to get the final trial solution, hence you need to multiply y twice by x such that:

`y = x^2(d_1x*e^x + 4d_2)`

`` You need to substitute the final trial solution for y in differential equation such that:

`y''-2y'+y = (x^2(d_1x*e^x + 4d_2))" - 2(x^2(d_1x*e^x + 4d_2))' + x^2(d_1x*e^x + 4d_2)`

`y''-2y'+y = (2x(d_1x*e^x + 4d_2)+x^2(d_1e^x +d_1x*e^x + 4d_2 ))' - 2(2x(d_1x*e^x + 4d_2) + x^2(d_1e^x +d_1x*e^x + ` `4d_2)) + x^2(d_1x*e^x + 4d_2) y''-2y'+y = 2(d_1x*e^x + 4d_2)+2x(d_1e^x + d_1x*e^x)+2x(d_1e^x +d_1x*e^x + 4d_2) + x^2(d_1e^x + d_1e^x + d_1x*e^x) - 4x(d_1x*e^x + 4d_2)- 2x^2(d_1e^x +d_1x*e^x + 4d_2)) + x^2(d_1x*e^x + 4d_2)`

`y''-2y'+y = 2d_1x*e^x + 8d_2 +2xd_1e^x + 2xd_1x*e^x + 2xd_1e^x + 2x^2d_1e^x + 8xd_2 + 2x^2d_1e^x + x^3d_1*e^x - 4x^2d_1e^x - 16xd_2 - 2x^2d_1e^x - 2x^3d_1e^x - 8x^2d_2 + x^3d_1e^x + 4x^2d_2`

`y''-2y'+y = x^2(- 2d_1e^x - 4d_2) + x(6d_1e^x- 8d_2)`

You need to equate `xe^x +4` and `x^2(- 2d_1e^x - 4d_2) + x(6d_1e^x- 8d_2)` such that:

`xe^x +4 = x^2(- 2d_1e^x - 4d_2) + x(6d_1e^x- 8d_2)`

Equating the coefficients of like powers yields:

`-2d_1e^x - 4d_2 = 0 =gt d_1e^x = 2d_2 =gt d_2 = (d_1e^x)/2`

`6d_1e^x- 8d_2 = e^x =gt 6d_1e^x- 4d_1e^x = e^x =gt 2d_1e^x = e^x =gt 2d_1 = 1 =gt d_1 = 1/2` `d_2 = e^x/4`

**Hence, the particular solution to the differential equation is `y_p = x^2e^x(x/2 + 1).` **