Prove $y=\sin(t^2)$ cannot be a solution on an interval containing $t=0$ of an equation $y'' + p(t)y'+q(t)y=0$

Question

I want to prove that $y=\sin(t^2)$ cannot be a solution on an interval containing $t=0$ of an equation $$y'' + p(t)y'+q(t)y=0$$ using the Wronskian and Abel's formula.

Let $y_{1}$ and $y_{2}=\sin(t^2)$ where $~y_{1}$ $\neq$ $y_{2}$ be linearly independent.

The Wronskian, $$W(y_{1},y_{2})(0) = 0~,$$ but how does this contradicts Abel's formula?

I am stuck for a long time, please help me.

Answer 1

First, you can use the following general property of linear equations or systems. For Wronskian of any system of solutions only one statement is true:

1. Wronskian is always zero, for all $x$ where solutions are defined. Thus, solutions of this system are linearly dependent.
2. Wronskian is always non-zero and solutions are linearly independent.

This general property can be proven without Abel's formula, but we can use it here (formulas are taken from here):

Theorem
Let $y_1$ and $y_2$ be any two solutions of $y'' + p ( x ) y'+ q ( x ) y =0$ , then $$W \lbrack y_1 ,y_2 \rbrack(x) = C \cdot \exp{\left (− \int\limits_{x_0}^{x} p ( u )\, du \right )},$$ where $C$ is a constant.

Exponential function never turns zero at any finite value, so this justifies our dichotomy: if $W[y_1, y_2](x) =0$ at some point then $C = 0$ and $W[y_1, y_2](x) \equiv 0$. I implicitly assume here that $p(x)$ is something like continuous function or at least integrable — this excludes the case when integral inside of exponential function is infinite.

Returning back to your question, we can state the following: were the function $\sin t^2$ a solution of second order ODE, then it must exist second function which is also a solution, it is linearly independent from $\sin t^2$ and thus their Wronskian must be everywhere non-zero. But we see that Wronskian of any function and $\sin t^2$ is always zero at $x = 0$. This is a contradiction which ends proof of your statement.

Answer 2

If $y(t) = \sin\left(t^2\right)$ then $$ y''(t) = 2 \cos \left(t^2\right) - (2t)^2 \sin \left(t^2\right) $$ hence $y''(0) \ne 0$ while $y'(0)=y(0)=0$.

Thus, for any functions $p$ and $q$, $y''(0)+p(0)y'(0)+q(0)y(0)\ne0$, which shows that $y$ solves no such differential equation.