ok, for 2nd order differential equations, if we have repeat roots, like for example y" +4y = cos2x(adsbygoogle = window.adsbygoogle || []).push({});

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

so how do we try for the particular solution?

i tried Acos2x + Bsin2x, it all cancelled out.

i tried (Ax+B)cos2x + (Cx+D)sin2x it all cancelled out too

then i tried just (Ax^{2}+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!

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# Second order DE involving repeat roots

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