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**iCeCuBEz v2** · 04-22-2013, 10:01 PM

It is known that the general equation of a homogeneous equation with constant coefficients that has an unrepeated pair of complex conjugate roots a plus or minus bi (with b not equal to 0) is as follows

e^ax(c1cosbx+c2sinbx)=y(x)

an example being

y''-4y'+5y=0

which has the characteristic equation

r^2-4r+5

which equals (by completing the square)

(r-2)^2+1=0

then solve for r in that characteristic equation yielding the complex conjugate roots 2 plus or minus i

then the general solution of the equation is

y(x)=e^(2x)(c1cosx+c2sinx)

I get how to do the procedure I just want to know why the general solution ends up being of the form e^ax(c1cosbx+c2sinbx)

04-22-2013, 10:01 PM	#1
iCeCuBEz v2 XFD Join Date: Mar 2008 Location: Connecticut Age: 33 Posts: 4,924	Differential Equations [a mere curiosity] It is known that the general equation of a homogeneous equation with constant coefficients that has an unrepeated pair of complex conjugate roots a plus or minus bi (with b not equal to 0) is as follows e^ax(c1cosbx+c2sinbx)=y(x) an example being y''-4y'+5y=0 which has the characteristic equation r^2-4r+5 which equals (by completing the square) (r-2)^2+1=0 then solve for r in that characteristic equation yielding the complex conjugate roots 2 plus or minus i then the general solution of the equation is y(x)=e^(2x)(c1cosx+c2sinx) I get how to do the procedure I just want to know why the general solution ends up being of the form e^ax(c1cosbx+c2sinbx)