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

Theorem

Let $\map {y_p} x$ be a particular solution to the homogeneous linear second order ODE:

$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$

on a closed interval $\closedint a b$.

Let there exist $\xi \in \closedint a b$ such that the curve in the cartesian plane described by $y = \map {y_p} x$ is tangent to the $x$-axis at $\xi$.

Then $\map {y_p} x$ is the zero constant function:

$\forall x \in \closedint a b: \map {y_p} x = 0$

Proof

Aiming for a contradiction, 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 $\xi$:

$\map {y_p} \xi = 0$

$\map { {y_p}'} \xi = 0$

Taking the Wronskian of $y_p$ and $y_2$:

$\map W {y_p, y_2} = y_p {y_2}' - {y_p}' y_2$

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.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.14$: Problem $6$