Find the homogeneous equation with constant coefficients of least order that has the given function as a solution.

y=3xe^-x + e^-xcos2x + 1

Question

Find the homogeneous equation with constant coefficients of least order that has the given function as a solution.

y=3xe^-x + e^-xcos2x + 1

experimentx · Answer

differentiate it couple of times and try to things on the right.

anonymous · Answer

well in order to get the homogeneous equation, dont all i need are the roots? and from 3xe^-1, that means it has to be a double root of -1 right... so (r+1)^2 and then from the constant at the end you get r=0, and then from the cos function you would get imaginary roots of -1+/-2i, so that would be (r^2+2r+5)

so (r^2+2r+5)(r+1)(r+1)r

y'''''+4y''''+11y'''+12y''+5y'

but to the 5th order doesnt seem like it would be to the lowest because there is no sin function in there nor is there e^-x in there either

experimentx · Answer

yeah ... seems okay

anonymous · Answer

is there a reason y those two were not included in the characteristic equation? or am i missing something

experimentx · Answer

what do you mean by those two?

anonymous · Answer

well to me... if i am right and those are the 5 roots then the function would look more like

y= C1e^-x + 3xe^-x + e^-xcos2x + e^-xsin2x + 1

and not just y=3xe^-x + e^-xcos2x +1

experimentx · Answer

yes ... the general solution of that DE should look like this one ... but you are not looking for general solution ... you are looking for particular solutions. also the general solution should be like this
y= C1e^-x + C2xe^-x + C3e^-xcos2x + C4e^-xsin2x + C5
...
for your case, just put C1 = 0, C2 = 3 ... and so on

anonymous · Answer

ohhhh duhhhhhh to me huh... i knew that... thank you for the help.

experimentx · Answer

yw