Solution of Linear 2nd Order ODE Tangent to X-Axis

Theorem
Let $y_p \left({x}\right)$ be a particular solution to the homogeneous linear second order ODE:
 * $(1): \quad \dfrac {\mathrm d^2 y} {\mathrm d x^2} + P \left({x}\right) \dfrac {\mathrm d y} {\mathrm d x} + Q \left({x}\right) y = 0$

on a closed interval $\left[{a \,.\,.\, b}\right]$.

Let there exist $\xi \in \left[{a \,.\,.\, b}\right]$ such that the curve in the cartesian plane described by $y = y_p \left({x}\right)$ is tangent to the $x$-axis at $\xi$.

Then $y_p \left({x}\right)$ is the zero constant function:
 * $\forall x \in \left[{a \,.\,.\, b}\right]: y_p \left({x}\right) = 0$

Proof
Suppose $y_p$ is not the zero constant function.

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another, there exists another particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent.

At the point $\xi$:


 * $y_p \left({\xi}\right) = 0$
 * ${y_p}' \left({\xi}\right) = 0$

Taking the Wronskian of $y_p$ and $y_2$:
 * $W \left({y_p, y_2}\right) = y_p {y_2}' - {y_p}' y_2$

But at $\xi$ this works out as zero.

It follows from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent that $y_p$ and $y_2$ cannot be linearly independent after all.

From this contradiction, $y_p$ must the zero constant function.