Particular integral in complementary function
Why does multiplying by $x$ work here? Is this a general rule when the particular integral in complementary function
2 Answers
$\begingroup$$$y''-5y'+6y=e^{2x}$$The characteristic polynomial is:$$r^2-5r+6 =0 \implies (r-2)(r-3)=0$$$$ \implies S_r= \{2,3\}$$The solution of the homogeneous equation is :$$y(x)=c_1e^{2x}+c_2e^{3x}$$So the particular solution should be $$y_p(x)=Axe^{2x}$$Normally the guess should be $Ae^{2x}$. But since $e^{2x}$ is already solution of the homogeneous equation, you need to multiply by $x$ the guess.
EDIT
A good exercice is to solve the following equation :$$y''-4y'+4y=e^{2x}$$Try it ..
$\endgroup$ 5 $\begingroup$The intrinsic structure of this phenomenon can be seen via the operator viewpoint:
If ${D_x}^2(y) - 5·D_x(y) + 6 y = e^{2x}$ then:
$( {D_x}^2 - 5·D_x + 6 )(y) = e^{2x}$.
$(D_x-2)( {D_x}^2 - 5·D_x + 6 )(y) = (D_x-2)(e^{2x})$.
$(D_x-2)(D_x-2)(D_x-3)(y) = 0$.
It should be clear that the extra polynomial factor arises from the multiplicity of $(D_x-2)$.
Of course, there are other details to fill in, but the intrinsic structure is here. An exactly analogous phenomenon happens with the forward difference operator for the same reason.
$\endgroup$ 1
