# 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?

• 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. Commented Feb 21, 2017 at 15:10
• u misread my question. $y_{1}$ $\neq$ $y_{2}$ Commented Feb 21, 2017 at 15:28
• do you mean en.wikipedia.org/wiki/Abel's_theorem for sequences? Commented Feb 21, 2017 at 20:17
• Abel's formula for Wronskian of solutions of 2nd order linear homogeneous DE Commented Feb 22, 2017 at 2:55

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.

• 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? Commented Feb 22, 2017 at 12:12
• 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. Commented Feb 22, 2017 at 12:25
• The Wronskian property is mentioned in my course. That is why i find it strange to use Abel's theorem in such a way. Commented Feb 22, 2017 at 14:16
• Frankly speaking me too :) Commented Feb 22, 2017 at 16:33
• i have more questions regarding ordinary differential equations, do you mind if i contact you about the questions? Commented Feb 23, 2017 at 4:59

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.

• 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. Commented Feb 21, 2017 at 15:32
• @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?
– Did
Commented Feb 22, 2017 at 8:07
• @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. Commented Feb 23, 2017 at 7:57
• @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. Commented Feb 23, 2017 at 10:57

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.