# Find the form of a particular solution to the differential equation. (Do not attempt to find the coefficients.)4y''-4y'+y=x( e^(x/2) + e^(-x/2) )+x^2

cosinusix | College Teacher | (Level 3) Assistant Educator

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First look at the characteristic equation of the homogeneous problem

`4r^2-4r+y=0 `

`r=1/2` is a double root.

A particular solution will be the sum of a particular solution of

`4y''-4y'+y=x^2`

plus a particular solution of  `4y''-4y'+y=xe^(-x/2)`

plus a particular solution of  `4y''-4y'+y=xe^(x/2)`

First equation: The second part is a polynomial function of degree 2. A particular solution will be a polynmial function of the same degree.

Second equation: the second part is the product of a polynomial function of degree 1 and an exponential function `e^(-x/2)`

-1/2 is not solution of the caracteristic equation.

A particular solution will be the product of polynomial function of the same degree and e^(-x/2)

Third equation: the second part is the product of a polynomial function of degree 1 and an exponential function `e^(x/2) `

1/2 is double solution of the caracteristic equation.

A particular solution will be the product of a polynomial function of degree 3 ( add 2) and `e^(x/2).`

In conclusion a particular solution will be in the form

`(ax^3+bx^2+cx+d)e^(x/2)+(fx+g)e^(-x/2)+hx^2+ix+j`

where the coefficients `a, b, c, d, f, g, h, i, j ` have to be determined.