Question about conditions required for an alternate method of solving reducible 2nd order ODEs.

We start with the second-order inhomogeneous ODE:

$$ y''(t) + h(t)y'(t) + q(t)y(t) = f(t), $$

and assume that $\phi_1(t)$ is a solution of the corresponding homogeneous equation:

$$ \phi_1''(t) + h(t)\phi_1'(t) + q(t)\phi_1(t) = 0. $$

Let $\phi_2(t) = v(t)\phi_1(t)$ be the second solution of the homogeneous equation, where $v(t)$ is to be determined. Substituting $\phi_2(t)$ into the homogeneous equation gives:

$$ \frac{v''}{v'} + \left[2\frac{\phi_1'(t)}{\phi_1(t)} + h(t)\right] = 0. $$

Now, for the inhomogeneous equation, we use variation of parameters. Let $v_1(t)$ and $v_2(t)$ be functions such that:

$$ v_2'(t) = \frac{f(t)}{W(t)}\phi_1(t), \quad v_1'(t) = -\frac{f(t)}{W(t)}\phi_2(t), $$

where $W(t)$ is the Wronskian of $\phi_1(t)$ and $\phi_2(t)$. From the relation $v_1'(t) = -v(t)v_2'(t)$, we find:

$$ v_1(t) = \int -v(t)v_2'(t) \, dt. $$

The particular solution of the inhomogeneous equation is then given by:

$$ y_p(t) = v_1(t)\phi_1(t) + v_2(t)\phi_2(t). $$

Substituting $v_1(t)$ and $v_2(t)$, we write:

$$ y_p(t) = v_2(t)\phi_2(t) - \phi_1(t)\int v(t)v_2'(t) \, dt. $$

Using $\phi_2(t) = v(t)\phi_1(t)$, this simplifies to:

$$ y_p(t) = \phi_1(t)\int v'(t)v_2(t) \, dt. $$

Next, substitute $v_2(t) = \int \frac{f(\sigma)}{W(\sigma)}\phi_1(\sigma) \, d\sigma$ into the expression for $y_p(t)$:

$$ y_p(t) = \phi_1(t) \int v'(t) \int \frac{f(\sigma)}{W(\sigma)}\phi_1(\sigma) \, d\sigma \, dt. $$

Using the relation $W(\sigma)/\phi_1(\sigma) = v'(\sigma)\phi_1(\sigma)$, we further simplify:

$$ y_p(t) = \phi_1(t) \int v'(t) \int \frac{f(\sigma)}{v'(\sigma)\phi_1(\sigma)} \, d\sigma \, dt. $$

Rewriting the nested integrals using $\theta$ and $\sigma$ for clarity, the particular solution becomes:

$$ y_p(t) = \phi_1(t) \int_t \exp\left(-\int_\sigma \left[2\frac{\phi_1'(s)}{\phi_1(s)} + h(s)\right] ds\right) \left(\int_\sigma \frac{f(\theta)}{\phi_1(\theta)}\exp\left(\int_\theta \left[2\frac{\phi_1'(u)}{\phi_1(u)} + h(u)\right] du\right) d\theta\right) d\sigma. $$

This matches the desired form. The solution is dependent only on $\phi_1(t)$, $f(t)$, and the coefficients of the differential equation, expressed as a double-integral where $\phi_1(t)$ and $f(t)$ are explicitly involved.

In deriving the solution, we assumed that $\phi_1(t)$ is a well-behaved solution to the homogeneous equation, satisfying $\phi_1(t) \neq 0$ on the domain of interest. This ensures that the Wronskian $W(t)$ is non-zero, allowing us to construct $\phi_2(t)$ as $v(t)\phi_1(t)$ and proceed with variation of parameters. Additionally, we assumed that the coefficients $h(t)$, $q(t)$, and the inhomogeneous term $f(t)$ are continuous, which guarantees the applicability of the method.

The use of $W(t)/\phi_1(t) = v'(t)\phi_1(t)$ is valid as it follows directly from the definition of the Wronskian and the form $\phi_2 = v\phi_1$. Furthermore, the final nested integral form relies on the convergence of these integrals, which holds under the assumption that $f(t)$ and $\phi_1(t)$ are sufficiently smooth and well-behaved.