Second order DE involving repeat roots

[quietrain]

ok, for 2nd order differential equations, if we have repeat roots, like for example y" +4y = cos2x

, we would have repeat roots +2i,-2i.

so how do we try for the particular solution?

i tried Acos2x + Bsin2x, it all canceled out.
i tried (Ax+B)cos2x + (Cx+D)sin2x it all canceled out too

then i tried just (Ax²+Bx)cos2x, but apparently that is wrong.

apparently we have to substitute y = Re(Z), and let z" +4z = e^i2x to solve

so my question is,

1)how do we know when we have to substitute, and when we can try AcosX +BsinX, also, for repeat roots, is the only way to solve is to let it be the real part of some substituted equation?

2)when the RHS is only cos 2x, do we have to try Acos2x +Bsin2x, or is Acos2x suffice? why?

help appreciated!

 

[LCKurtz]

Try Axsin(2x) + Bxcos(2x). It will work.

And your effort with (Ax+B)cos2x + (Cx+D)sin2x should have worked too.

 

[quietrain]

oh isee. i just have to assume my B and D take on the values of 0? for convenience?

thanks!

 

[LCKurtz]

oh isee. i just have to assume my B and D take on the values of 0? for convenience?

thanks!

It isn't a matter of convenience. If you equate like terms on both sides after you substitute into the equation, those values must be 0 to make it work. In other words, if you don't have an x cosx on the right side you can't have one on the left, etc.

 

[quietrain]

oh i see ha.. ok thanks a lot!