Let the homogeneous coefficients become functions

Try

Suppose y₁ and y₂ form an independent homogeneous basis for y″+p(t)y′+q(t)y=0. A forced solution might be built as yₚ=u₁(t)y₁(t)+u₂(t)y₂(t), allowing the coefficients to vary with time. Two unknown coefficient functions appear for one desired function, so we can impose an additional convenient constraint.

Choose u₁′y₁+u₂′y₂=0. This cancels the extra terms in the first derivative, leaving yₚ′=u₁y₁′+u₂y₂′. Write out the product rule to see exactly what has been removed.

Predict

Differentiate once more and substitute into the forced equation y″+py′+qy=g. Predict which terms vanish because y₁ and y₂ already solve the homogeneous equation. The remaining equation should involve u₁′ and u₂′, not second derivatives of those coefficients.

Experiment

The two equations for the coefficient derivatives are u₁′y₁+u₂′y₂=0 and u₁′y₁′+u₂′y₂′=g. Their determinant is the Wronskian W=y₁y₂′−y₁′y₂. Solving gives u₁′=−y₂g/W and u₂′=y₁g/W.

For y″+y=sec t on the interval from −π/2 to π/2, use y₁=cos t and y₂=sin t. Here W=1 and cos t is positive.

Coefficient derivative	Simplification	One antiderivative
u₁′=−sin t sec t	−tan t	ln(cos t)
u₂′=cos t sec t	1	t

This yields yₚ=cos t ln(cos t)+t sin t. Check the sign by differentiating ln(cos t), which is −tan t. A sign error here would reverse part of the forcing.

Understand

The method requires a normalized equation with leading coefficient one. If the original equation is a(t)y″+b(t)y′+c(t)y=f(t), divide by a only on an interval where a is nonzero, then use g=f/a. Forgetting this normalization changes the coefficient system and the forcing.

For continuous p,q,g on an interval and an independent homogeneous pair, the Wronskian remains nonzero throughout that interval. Integrating u₁′ and u₂′ then constructs a particular solution. Constants introduced in these integrals multiply y₁ and y₂, so they are already part of the homogeneous family and may be set to zero when seeking just one particular solution.

Verify the example explicitly. The first derivative of cos t ln(cos t) is −sin t ln(cos t)−sin t. Adding the derivative of t sin t cancels the isolated sine terms, leaving yₚ′=−sin t ln(cos t)+t cos t. Differentiating again and adding yₚ leaves sin²t/cos t+cos t=sec t. The interval restriction keeps the forcing and logarithm well-defined.

This method can leave unevaluated definite integrals. That is still a valid mathematical representation, and numerical quadrature may evaluate it. Lack of an elementary antiderivative is not lack of a solution.

Apply

Calculate W for e^t and e^(−t): it is −2, not zero. Then use a simple forcing to compute the two coefficient derivatives and compare with an undetermined-coefficients solution. The formulas can differ by a homogeneous term while representing the same full family.

In your written work, identify the interval, normalized forcing, basis, Wronskian, coefficient system, and final substitution. This sequence makes the assumptions visible and provides several independent opportunities to catch errors.

Continue

Next interpret second-order equations as mechanical models. Energy and damping provide qualitative checks on the formulas and help distinguish transient motion from a persistent forced response.