Find the form a particular solution to the differential equation. I do not attempt to find the coefficients. y''+6'+13y=x^3+xe^(-3x)+(e^(-3x)+1)sin(2x)

1 Answer | Add Yours

Top Answer

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should come up with the substitution:

`y_p = x^3+xe^(-3x)+(e^(-3x)+1)sin(2x)`

Hence, `y_p''+ 13y_p + 6y_p' = (x^3+xe^(-3x)+(e^(-3x)+1)sin(2x))" + 13(x^3+xe^(-3x)+(e^(-3x)+1)sin(2x)) + 6(x^3+xe^(-3x)+(e^(-3x)+1)sin(2x))'`

You need to differentiate `y_p`  with respect to x such that:

`y_p' = 3x^2 + e^(-3x) - 3xe^(-3x)-3e^(-3x)*sin 2x + 2(e^(-3x)+1)cos(2x)`

You need to differentiate `y_p"`  with respect to x such that:

`y_p'' = 6x - 3e^(-3x) + 9xe^(-3x) + 9e^(-3x)sin 2x - 6e^(-3x)*cos 2x- 6e^(-3x)*cos 2x - 4(e^(-3x)+1)sin(2x)`

`y_p'' = 6x - 9e^(-3x) + 9xe^(-3x) + 5e^(-3x)sin 2x -12e^(-3x)*cos 2x - 4sin 2x` `y_p''+ 13y_p + 6y_p' =6x - 9e^(-3x) + 9xe^(-3x) + 5e^(-3x)sin 2x -12e^(-3x)*cos 2x - 4sin 2x + 13x^3 + 13xe^(-3x) + 13e^(-3x)sin(2x) + 13 sin 2x + 18x^2 + 6e^(-3x) - 18xe^(-3x)-18e^(-3x)*sin 2x + 12e^(-3x)cos(2x) + 6cos 2x`

Reducing opposite terms yields:

`y_p''+ 13y_p + 6y_p' =13x^3 + 18x^2 + 6x- 3e^(-3x) + 4xe^(-3x) + 9sin 2x + 6cos 2x`

You should come up with the substitution:

`y_p = Ax^3+Bx^2 + Cx + De^(-3x) + Ee^(-3x)sin(2x) + Fsin 2x`

Hence, equating both forms of `y_p,Ax^3+Bx^2 + Cx + De^(-3x) + Ee^(-3x)sin(2x) + Fsin 2x = 13x^3 + 18x^2 + 6x- 3e^(-3x) + 4xe^(-3x) + 9sin 2x + 6cos 2x ` yields A,B,C,D,E,F.

We’ve answered 318,990 questions. We can answer yours, too.

Ask a question