Solving DE by undetermined Coefficients

[helpppmeee]

Homework Statement

y'' + y = xsinx where y(0) = 0 and y'(0) = 1

Homework Equations

yc = C1sinx + C2cosx

The Attempt at a Solution

So my attempt includes the yc which undoubtedly is correct. I now try to solve yp

yp = Axsinx + Bxcosx
y''p = A(2cosx - xsinx) + B(-2sinx - xcosx)

Substituting: y''p + yp = xsinx, clearly doesn't equal xsinx. There are no comparable values to xsinx to even try and solve for A or B.

Attempting using a different algorithm: yp = (A + Ax)sinx + (B + Bx)cosx
Notice overlap from yp to yc, therefore multiply by x: yp = (Ax +Ax^2)sinx + (Bx + Bx^2)cosx
y''p = A(2cosx + 2sinx -xsinx +4xcosx -x^2sinx) + B(-2sinx + 2cosx - 4xsinx - x^2cosx)

and y''p + yp = A(2cosx + 2sinx + 4xcosx + xsinx) + B(-2sinx + 2cosx - 4xsinx + xcosx)

Arrgggh this is really troubling me. I honestly have no idea what I'm doing wrong or how to proceed with the Initial Value Problem afterwards.

 

[SammyS]

Homework Statement

y'' + y = xsinx where y(0) = 0 and y'(0) = 1

Homework Equations

yc = C1sinx + C2cosx

The Attempt at a Solution

So my attempt includes the yc which undoubtedly is correct. I now try to solve yp

yp = Axsinx + Bxcosx
y''p = A(2cosx - xsinx) + B(-2sinx - xcosx)

Substituting: y''p + yp = xsinx, clearly doesn't equal xsinx. There are no comparable values to xsinx to even try and solve for A or B.

Attempting using a different algorithm: yp = (A + Ax)sinx + (B + Bx)cosx
Notice overlap from yp to yc, therefore multiply by x: yp = (Ax +Ax^2)sinx + (Bx + Bx^2)cosx
y''p = A(2cosx + 2sinx -xsinx +4xcosx -x^2sinx) + B(-2sinx + 2cosx - 4xsinx - x^2cosx)

and y''p + yp = A(2cosx + 2sinx + 4xcosx + xsinx) + B(-2sinx + 2cosx - 4xsinx + xcosx)

Arrgggh this is really troubling me. I honestly have no idea what I'm doing wrong or how to proceed with the Initial Value Problem afterwards.

Try some general linear function times sin(x) & cos(x).

$\displaystyle y_p=(Ax+B)\sin(x)+(Cx+D)\cos(x)$

 

[helpppmeee]

OH my, rookie mistake LOL. but still, i only get that y''p + yp = 2Acosx - 2Csinx. My logic is telling me to try out yp = (Ax^2 + Bx + C)sinx + (Dx^2 + Ex + F)cosx, based upon my classroom notes. I try this and get
y''p + yp = 2Asinx + 2Dcosx + 2(2Ax+B)cosx -2(2Dx + E)sinx = xsinx

now i understand that 2(2Ax+B)cosx -2(2Dx + E)sinx = xsinx and so 2(2Ax+B)cosx = 0 and -2(2Dx + E)sinx = xsinx
as well as: 2Asinx + 2Dcosx = 0, but this seems like an unsolvable puzzle to me. I also believe i may be overthinking things A LOT so um yeah. What's going on here?

 

[helpppmeee]

Sorry but i edited my post and am not sure if you can see that from outside the thread so... bump

 

[vela]

Your initial guess at the particular solution would normally be $y_p = (Ax+B)\sin x + (Cx+D)\cos x$; however, because this isn't linearly independent from the homogeneous solution, you need to multiply it by the lowest power of $x$ to make it independent.