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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 $y_{2}=\sin(t^2)$, $y_{1}$ $\neq$ $y_{2}$ 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 :(

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    Well of course $y_1$ and $y_2$ are going to be linearly dependent, they are the same function. Independence tests whether or not you can write one in terms of the other. Since $\sin(t^2) = \sin(t^2)$, they are linearly dependent. – 2017-02-21
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    u misread my question. $y_{1}$ $\neq$ $y_{2}$ – 2017-02-21
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    do you mean https://en.wikipedia.org/wiki/Abel's_theorem for sequences? – 2017-02-21
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    Abel's formula for Wronskian of solutions of 2nd order linear homogeneous DE – 2017-02-22

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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.

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    Yes i thought of this yesterday. I feel that it is a contradiction to the assumption that the solutions $\sin t^{2} $ and $y_{1}$ are linearly independent, instead of contradiction to the Abel's theorem. What do you feel about this? – 2017-02-22
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    I agree with you. I think that here Abel's theorem is a sort of "cheap" substitution for this Wronskian property that I've mentioned in the beginning of my answer. It might be the reason to use Abel's theorem if this property wasn't mentioned in your course. – 2017-02-22
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    The Wronskian property is mentioned in my course. That is why i find it strange to use Abel's theorem in such a way. – 2017-02-22
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    Frankly speaking me too :) – 2017-02-22
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    i have more questions regarding ordinary differential equations, do you mind if i contact you about the questions? – 2017-02-23
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    I mostly follow only ODE and dynamical systems questions, so if you ask something on MSE, I will likely notice the question and try to come up with an idea at least :) – 2017-02-23
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    I have a question, can you help me? http://math.stackexchange.com/questions/2157986/behaviour-of-2nd-order-homogeneous-linear-ode-with-constant-coefficients – 2017-02-24
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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.

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    Your method shows the LHS = 2 $\neq$ RHS, which complete the proof. But i'm asking for a proof using the method mentioned in the post. – 2017-02-21
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    @LittleRookie Any solution based on the Wronskian will use, in a more or less disguised way, the simple fact that $y(0)=y'(0)=0\ne y''(0)$ for $y(t)=\sin(t^2)$... hence why not prefer the argument which goes to the heart of the matter? – 2017-02-22
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    @Did I know both methods, but according to my course material, it mentioned about contradicting the Abel's formula, which i could not understand why. Hence i post this question, asking how it may contradict the Abel's formula. – 2017-02-23
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    @Did Wronskian and linear-(in)dependence based methods don't go for the second derivative here :) However I like the brevity of the method in this answer. – 2017-02-23