ODE (solve for particular integral) am i right?

[soonsoon88]

ODE (solve for particular integral) am i right??

Homework Statement

That is all the examples that i know to determine the particular solution.
are all of them correct? Note: e= exponent

a)x^3 ==> Ax^3 + Bx^2 + Cx + D
b)xe^2 ==> e^2(Ax +B)
c)sin x ==> Asin x + Bcos x
d)cos x ==> Asin x + Bcos x
e)sin x + cos x ==> Asin x + Bcos x + Csin x + Dcos x
f)sin x + xcos x ==> Asin x + Bcos x + (Cx+D)sin x + (Ex+F)cos x
g)x^2(sin x) ==> (Ax+B)sin x + (Cx+D)cos x
h)e^x (sin x) ==> e^x( Asin x + Bcos x)

Homework Equations

Am i right?
is it still got another type of particular?
that is all i know only.

The Attempt at a Solution

 

[jimmypoopins]

that looks about right, except for e) and g). also, on b, it it supposed to be xe^(2x)?

on e), if you factor your answer you get
Asin x + Bcos x + Csin x + Dcos x = sinx(A+C) + cosx(B+D). since A, B, C, D are just constants, you can tell that you don't really need the C or the D.

on g), you're missing an x^2 term.

 

[HallsofIvy]

One caveat: If a function of that form is already a solution to the associated homogeneous equation, then you will need to multiply by 4.

For example, if the problem is y"- 5y'+ 6y= e^x, so the characteristic equation is r²- 5r+ 6= (r- 3)(r- 2)= 0, which has roots 2 and 3, so that the solutions to the homogeneous equation are e^2x and e^3x, then, yes, I would try a particular solution of the form y= Ae^x.

But if the problem is y"- 4y'+ 3y= e^x, so the characteristic equationj is r²- 4r+ 3= (r-1)(r- 3)= 0, which has roots 1 and 3, so that the solutions to the homogeneous equation are e^x and e^3x, then Ae^x will give 0, not e^x, for any A. Now I would try y= Axe^x.

If the problem is y"- 2y'+ y= e^x, so that the characteristic equation is r²- 2r+ 1= (r-1)²= 0, which has 1 as double root, so that the solutions to the homogeneous equation are e^x and xe^x, then I would have to tr Ax²e^x.