Given $y_1(x)=\sin(x^2)$ and $y_2(x)=\cos(x^2)$, I constructed a linear, homogenic ODE of order 2 by solving: $$ \begin{vmatrix} y & y_1 & y_2 \\ y' & y_1' & y_2' \\ y'' & y_1'' & y_2'' \\ \end{vmatrix}=0 $$

Now, I noticed that the Wronskian of $y_1$ and $y_2$ at $x=0$ equals $0$. But the Wronskian of independent solutions is never $0$. And if it $0$ at one point, it is zero everywhere, which I don't see happening here, as

$$ W(y_1,y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \\ \end{vmatrix}=-2x $$

How does this make sense?


make sure that the coefficient of the highest order derivative doesn't vanish anywhere in the domain and the coefficient of the rest of the derivatives are continuous.specially check at x=0

  • $\begingroup$ I agree. Indeed, at $x=0$ the coefficient of $y''$ is the determinant $$\det\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$$ $\endgroup$ Feb 8 '16 at 21:13
  • $\begingroup$ Thank you for your answer. If the coefficient of the highest order derivative vanishes, what does it imply? $\endgroup$
    – Whyka
    Feb 8 '16 at 21:52
  • $\begingroup$ See that doesn't imply anything straight away but you can't just blindfoldly apply the properties of wronskian without taking account the other assumptions under which the properties of wronskian get applied.please go through your concept and you yourself will find an answer for your question.... $\endgroup$
    – user311790
    Feb 9 '16 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.