The given differential equation for the spring-oscillation problem is:

$m \frac{d^2 y}{dt^2} + ky = A \cos(\alpha t) + B \sin(\alpha t)$

This is a second-order linear non-homogeneous differential equation, where $m > 0$ represents the mass, $k > 0$ represents the spring constant, and the external force is given by $f(t) = A \cos(\alpha t) + B \sin(\alpha t)$.

Understanding Resonance:

Resonance occurs when the frequency of the external force matches the natural frequency of the system. The natural frequency of the spring-mass system is given by:

$\omega_0 = \sqrt{\frac{k}{m}}$

For resonance to occur, the frequency of the external force must match this natural frequency. The external force has components $\cos(\alpha t)$ and $\sin(\alpha t)$ with frequency $\alpha$. Resonance occurs when:

$\alpha = \omega_0 = \sqrt{\frac{k}{m}}$

Solving the Differential Equation:

The complementary solution to the homogeneous part $m \frac{d^2 y}{dt^2} + ky = 0$ is:

$y_c(t) = C_1 \cos\left(\sqrt{\frac{k}{m}} t\right) + C_2 \sin\left(\sqrt{\frac{k}{m}} t\right)$

For the particular solution, since resonance is occurring ($\alpha = \sqrt{\frac{k}{m}}$), the standard approach leads to solutions involving a term proportional to $t$ (since the standard sinusoidal response is not sufficient in the resonant case).

The particular solution will be of the form:

$y_p(t) = t \left( A' \cos\left(\sqrt{\frac{k}{m}} t\right) + B' \sin\left(\sqrt{\frac{k}{m}} t\right) \right)$

Substituting this into the differential equation and matching coefficients with the external force term, we get:

[ y_p(t) = \frac{1}{2} \sqrt{\frac{m}{k}} \left( A t \sin\left(\sqrt{\frac{k}{m}} t\right) - B t \cos\left(\sqrt{\frac{k}{m}} t\right)