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The repeated-root constant-coefficient differential equation has characteristic polynomial $(r+2{)}^2$. Its two fundamental solutions areThey are linearly independent: their Wronskian is $e^{-4x}$, which never vanishes. For the forced equation, exploit the repeated factor by writing $y=e^{-2x}u$. Direct differentiation gives $(D+2{)}^{2}y=e^{-2x}{u}^{\prime \prime }$, so the forcing reduces to $u^{\prime \prime }=1$. The initial data imply $u(0)=0$ and $u^{\prime }(0)=0$, hence $u=x^2/2$. ThereforeThe extra factor $x^2$ reflects resonance of the forcing with the repeated characteristic root; the exponential substitution obtains it without guessing a particular solution.

Promote the canonical field and momentum $\pi =f^{\prime }$ to operators satisfying the canonical commutation relation, $[\hat{f}(\tau ,\mathbf{x}),\hat{\pi }(\tau ,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$. A real field has Fourier reality condition ${\hat{f}}_{\mathbf{k}}^{†}={\hat{f}}_{-\mathbf{k}}$. Its oscillator expansion isThe creation and annihilation operators add or remove excitations in the selected mode basis, withCanonical normalization requires the Wronskian $u_k{u}_k^{\ast \prime }-u_k^{\ast }{u}_k^{\prime }=i$.

For the Bunch-Davies vacuum, choose the positive-frequency short-wavelength behavior $u_k\sim e^{-ik\tau }/\sqrt{2k}$ as $k|\tau |\to \infty$. For the proposed solution, direct differentiation givesSubstitution cancels every term in $u_k^{\prime \prime }+(k^2-2/{\tau }^2)u_k$. Its Wronskian is $i$, and its large-$|k\tau |$ limit is the required flat-spacetime positive-frequency mode. ThusA term with the conjugate frequency is another allowed classical solution, but would describe a different quantum state, through a Bogoliubov transformation. Its coefficient is set to zero by the early-time vacuum condition, not by absence of a second solution to the differential equation.

Substitute $y_2=v{y}_1$ and use $y_1^{\prime \prime }+p{y}_1^{\prime }+q{y}_1=0$ to cancel the terms proportional to $v$. The remaining ordinary differential equation isOn an interval where $y_1\ne 0$, divide through and put $w=v^{\prime }$:Thus reduction of order givesA nonzero multiplicative constant gives an independent solution; adding a constant inside the integral adds a multiple of $y_1$. The Wronskian is proportional to $e^{-\int p}$ and is nonzero on the interval. The undivided equation remains meaningful at a zero of $y_1$, but this integral representation then needs continuation rather than direct division.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.