I'm trying to tackle the following question:

Let $\displaystyle y'+p(x)y=0$, such that $\displaystyle p(x)$ is continuously differentiable and bounded.

Show that every solution which is not $\displaystyle y\equiv 0$ is not intersecting with the x axis.

Given that $p(x)$ continuously differentiable implies that $f(x,y)=-p(x)y$ and $\displaystyle \frac{\partial{f(x,y)}}{\partial{y}}$ are continouos, i.e the equation has single solution to any initial condition.

Second, I tried to use integrating factor and got $\displaystyle y=Ce^{-\int{p(x)}\text{d}x}$, but I don't know how to continue? Why is it important the $\displaystyle |p(x)|\le M$?

Please help, thank you!

  • $\begingroup$ Hint: $\dfrac {y'}{y}=-p(x)$ means that $\log y$ is contiuosly differentiable and bounded. $\endgroup$ – Emilio Novati Mar 27 '15 at 10:56
  • $\begingroup$ @EmilioNovati, unfortunately I don't see how to continue :/ $\endgroup$ – Galc127 Mar 27 '15 at 11:03
  • $\begingroup$ If $y=0$ than $\log y$ does not exists ... $\endgroup$ – Emilio Novati Mar 27 '15 at 11:05
  • $\begingroup$ @EmilioNovati, thanks a lot! $\endgroup$ – Galc127 Mar 27 '15 at 11:23

Since $y\equiv 0$ is already a solution, by uniqueness the whole $x$-axis is excluded from all other solutions. Now to change sign you would have to cross the $x$-axis…

Or assume that for some solution $y(t^*)=0$, then $y$ also solves the IVP with $y(t^*)=0$ and the only solution of that IVP is $y\equiv 0$.

  • $\begingroup$ Thanks, your answer was very helpful. $\endgroup$ – Galc127 Mar 27 '15 at 11:24

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