second order Differential eqt

You have set the wrong particular solution. That's why you didn't get the correct answer.

d/dx(dy/dx) + 4y = sin2x

Auxiliary equation: k2 + 4 = 0

k = +- 2i

So, complementary solution:

yc = Asin2x + Bcos2x

Now, we should let the particular solution be Cxsin2x + Dxcos2x instead of Csin2x + Dcos2x, otherwise, it would be the complementary solution and it is meaningless to do so.

yp = Cxsin2x + Dxcos2x

dyp/dx = Csin2x + 2Cxcos2x + Dcos2x - 2Dxsin2x

= (C - 2Dx)sin2x + (2Cx + D)cos2x

d/dx(dyp/dx) = 2(C - 2Dx)cos2x - 2Dsin2x - 2(2Cx + D)sin2x + 2Ccos2x

= 4(C - Dx)cos2x - 4(Cx + D)sin2x

Putting back to the equation,

[4(C - Dx)cos2x - 4(Cx + D)sin2x] + 4(Cxsin2x + Dxcos2x) = sin2x

-4Dsin2x + 4Ccos2x = sin2x

Comparing coefficients,

D = -1/4, C = 0

Therefore, the general solution, y

= yc + yp

= Asin2x + Bcos2x - 1/4 xcos2x

= Asin2x + (B - 1/4 x)cos2x



2009-05-05 13:55:09 補充：
如果你的complementary solution是Csin2x + Dcos2x這個形式的話，你便要set particular solution為Cxsin2x + Dxcos2x

如果complementary solution為Csin2x + Dxsin2x + Ecos2x + Excos2x的話，你便要set p.s. 為Cx^2sin2x + Dx^2cos2x

如此類推。即set為polynomial高一個degree便可以了。

有咩情況係要set PI= Cxsin2x + Dxcos2x instead of Csin2x + Dcos2x?
我好多時都係就咁用 Csin2x + Dcos2x.
唔係好明白..