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Suppose $y_1$ and $y_2$ potentially form a fundamental set of solutions to a second order linear homogeneous ODE given by $y''+p(x)y'+g(x)y=0$. If $y_2(x_o)=y_1(x_0)=0$. The Wronskian at $x_o$ is then $W(y_1,y_2)(x_o)=y_1(x_o)y_2'(x_o)-y_2(x_o)y_1'(x_o)=0$.

The Wronskian is also given by Abel's theorem as $W(y_1,y_2)(x)=W(x_o)e^{-\int_{x_o}^x p(x)dx}$, hence $W(x)=0$ for all $x$ and therefore $y_1$ and $y_2$ cannot form a fundamental set of solutions.

Now consider the case $y_1=\sin(x)$, $y_2=\sin(2x)$. At $x=0$, they are both zero and so from the statements above we'd expect them to never form a fundamental set of solutions to a scond order linear homogeneous ODE. But they do, for example they are solutions to $\sin^2(x)y''-3\sin(x)\cos(x)y'+(1+2\cos^2(x))y=0$. How can this be?

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  • $\begingroup$ Your differential equation has discontinuous coefficients at zero once you put it in the form $y^{\prime\prime}+p(x)y^\prime+g(x)y=0$. $\endgroup$
    – Keenan Kidwell
    Nov 7, 2015 at 0:53
  • $\begingroup$ your equation have the following solution $$y=i|\sin x| \left(c_2 \cos x+c_1\right)$$ note the discontinuity $\endgroup$
    – Michael Medvinsky
    Nov 7, 2015 at 0:58

1 Answer 1

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The only think you are missing is that the condition of continuity for $p(x)$ and $q(x)$ for all $x_0$ belonging to $[a,b]$.

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