Given the following Cauchy-Euler equation
$$t^2y''-ty'-3y=4t^2+12,~t>0$$
one can first find the homogeneous roots through the characteristic equation
$$r^2-2t-3=0$$
$$(r-3)(r+1)=0$$
$$r_1=3,~r_2=-1$$
to find the homogeneous solution of
$$y_h(t)=c_1t^3+c_2t^{-1}$$
Next, one can find the particular solution by the method of undetermined coefficients. Since the right-hand side contains $4t^2+12$, a perfectly reasonable guess for the particular solution is
$$y_p(t)=At^2 + Bt+C \tag{1}$$
where upon differentiating twice and relating coefficients forms
$$y_p(t)=-\frac{4}{3}t^2-4$$
However, one is also able to guess
$$y_p(t)=At^2 + B \tag{2}$$
and arrive at the correct particular solution.
One explanation for this is because the right-hand side of the equation has powers with exponents $2$ and $0$, neither of which is a root of the characteristic polynomial. Since a multiple of $t$ is not present in the right-hand side of the equation, it isn't necessary to include a multiple of $t$ in the guess for the particular solution.
Is this logic correct? Would the same logic apply to any similar second order Cauchy-Euler equation? For example, suppose the original equation was changed to
$$t^2y''-ty'-3y=4t^4+4t^2+12,~t>0$$
then one could guess either
$$y_p(t)=At^4 + Bt^3+Ct^2+Dt+E \tag{3}$$
or
$$y_p(t)=At^4 + Bt^2+C \tag{4}$$
and still arrive at the correct particular solution of
$$y_p(t)=\frac{4}{5}t^4-\frac{4}{3}t^2-4$$
Is it unnecessary to include a power of $t$ in the guess if it isn't included in the right-hand side of the original equation?
Best Answer
In the equivalence $t=e^x$, $u(t)=y(t)=y(e^x)$, the powers $t^k$ on the Cauchy-Euler side play the same role as the exponentials $e^{kt}$ on the constant coefficients side. In the same way that you need no additional terms for pure exponentials that are not in resonance, you also need no additional terms for pure powers $t^k$ on the right side of the Euler-Cauchy equation (naming in dispute, usually Euler-Cauchy is homogeneous).
Or by contrast, to get additional terms you would need to find a coefficient in front of some $t^k$ that is a polynomial in $\ln t$, and then these additional terms would have to fill the powers of $\ln t$ in the method of undetermined coefficients.