Year: 1941
Topic: Differential Equations Analysis

Problem:
Find the differential equation satisfied by the product $z$ of any two linearly independent integrals of the equation\[ y'' + y'P(x) + yQ(x) = 0. \]

Solution:
Suppose $y_1$ and $y_2$ are two linearly independent solutions of the given differential equation. Then any two solutions have the form $u = ay_1 + by_2$ and $v = cy_1 + dy_2$. Since $uv$ falls in the linear space spanned by $y_1^2, y_1y_2, y_2^2$, we expect to find that it satisfies a linear differential equation of the third order. Letting $z = uv$ we have\[ \text{(1)} \quad u'' + Pu' + Qu = 0 \] \[ \text{(2)} \quad v'' + Pv' + Qv = 0 \] \[ \text{(3)} \quad z' = uv' + u'v \] \[ \text{(4)} \quad z'' = uv'' + 2u'v' + vu''. \]
We find\[ \text{(5)} \quad z'' + Pz' + 2Qz = u(v'' + Pv' + Qv) + v(u'' + Pu' + Qu) + 2u'v' = 2u'v'. \]
Differentiating (5), we get\[ \text{(6)} \quad z''' + Pz'' + (P' + 2Q)z' + 2Q'z = 2u'v' + 2v'u''. \] Next multiply (1) by $2v'$ and (2) by $2u'$ and add to obtain\[ \text{(7)} \quad 2u'v'' + 2v'u'' + 4Pu'v' + 2Qz' = 0. \] Multiply (5) by $2P$ to get\[ \text{(8)} \quad 2Pz'' + 2Pz' + 4PQz = 4Pu'v'. \] Add (6), (7), and (8), cancelling the terms that appear on the right, to get\[ \text{(9)} \quad \boxed{z''' + 3Pz'' + (2P^2 + P' + 4Q)z' + (4PQ + 2Q')z = 0}, \] a third-order differential equation satisfied by the product $z$ of any two solutions of the original equation.