How to solve it by the method of undetermined coefficient. $$(D^3-D^2+3D+5)y = e^x \cos x$$
As CF of it is $$c_1e^{-x} + e^x ( c_2 \cos2x + c_3 \sin2x)$$ and $e^x$ is common in both CF and RHS of equation but if we open $\sin$ and $\cos$ function of CF then $e^x \cos x$ is common in both CF and RHS of equation.
How can I solve it???
On
$$(D^3-D^2+3D+5)y = e^x \cos x$$ For the particular solution the guess should be: $$y_p=e^x(A \cos x +B \sin x)$$ Since the CF has trig function $\cos(2x),\sin(2x)$ this differs from the inhomogeneous part of the DE.
What do you mean to open trigonometric functions in CF ? Do you mean that $\cos (2x) = 2\cos ^2 x -1 $ ? and $\sin 2x = 2 \sin x \cos x$ ? If thats what you mean then these trigonometric forms are still different from the inhomogeneous part.
Try this link: https://en.wikipedia.org/wiki/Method_of_undetermined_coefficients
It provides specific examples at the middle and end.