**Proof plan.** Since $Y_1(x)$ and $Y_2(x)$ form a basis for the two-dimensional phase space at each $x$, we can write $Y_p(x) = u(x)Y_1(x) + v(x)Y_2(x)$ for [functions](/page/Function) $u, v$ to be determined. Matching both components of this vector equation, plus requiring compatibility of the first and second components, yields a $2 \times 2$ linear system for $u'$ and $v'$ involving the Wronskian. Solving this system and integrating gives the formula. **Step 1: Ansatz and compatibility.** Write $y_p = u(x)y_1 + v(x)y_2$. Differentiating: \begin{align*} y_p' = u'y_1 + uy_1' + v'y_2 + vy_2'. \end{align*} Impose the constraint $u'y_1 + v'y_2 = 0$ so that $y_p' = uy_1' + vy_2'$ (this is the compatibility condition ensuring the second component of the solution vector matches). **Step 2: Substitute into the ODE.** Differentiating again: $y_p'' = uy_1'' + u'y_1' + vy_2'' + v'y_2'$. Substituting into $y'' + py' + qy = f$: \begin{align*} u(y_1'' + py_1' + qy_1) + v(y_2'' + py_2' + qy_2) + u'y_1' + v'y_2' = f(x). \end{align*} Since $y_1$ and $y_2$ are complementary functions, the bracketed expressions vanish, leaving: \begin{align*} u'y_1' + v'y_2' = f(x). \end{align*} **Step 3: Solve the linear system.** Combining with the constraint from Step 1: \begin{align*} \begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix} \begin{pmatrix} u' \\ v' \end{pmatrix} = \begin{pmatrix} 0 \\ f \end{pmatrix}. \end{align*} The determinant of the coefficient matrix is the Wronskian $W = y_1 y_2' - y_2 y_1'$, which is non-zero by [Abel's Theorem](/theorems/834) (since $y_1, y_2$ are linearly independent). Solving by Cramer's rule: \begin{align*} u'(x) = -\frac{y_2(x) f(x)}{W(x)}, \qquad v'(x) = \frac{y_1(x) f(x)}{W(x)}. \end{align*} **Step 4: Integrate.** \begin{align*} y_p(x) = y_2(x) \int^x \frac{y_1(t) f(t)}{W(t)} \, dt - y_1(x) \int^x \frac{y_2(t) f(t)}{W(t)} \, dt. \end{align*} Changing the lower [limits](/page/Limit) of [integration](/page/Integral) simply adds a multiple of a complementary function, so $y_p$ remains a particular integral regardless of the choice of lower limit.