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


*

*Wronskian is always zero, for all $x$ where solutions are defined. Thus, solutions of this system are linearly dependent. 

*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.
A: Or seen another way, the basic facts in the answer of gt6989b (or just $\sin(t^2)=t^2+O(t^4)$) give the initial conditions 
$$y(0)=y'(0)=0~\text{ for } ~y''+py'+qy=0.$$ 
By the general theory of linear ODE, especially the uniqueness claims, $y$ must be the zero function. Which the given $y$ is not. 
