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Michael Renardy
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The first equation is solvable in closed form, and then the second equation becomes a Riccati equation. For that, you have closed form solutions only for special values of $\alpha$. Some general observations: $y$ remains bounded if and only if it is always nonnegative. A necessary condition for that is that $y(0)$ is positive and $x$ is integrable. Whether $x$ is integrable depends on $\alpha$. If $x$ is not integrable, then $y$ willcannot remain nonnegative. It may remain nonnegative if $x$ is integrable and $y(0)$ is large enough.

The first equation is solvable in closed form, and then the second equation becomes a Riccati equation. For that, you have closed form solutions only for special values of $\alpha$. Some general observations: $y$ remains bounded if and only if it is always nonnegative. A necessary condition for that is that $y(0)$ is positive and $x$ is integrable. Whether $x$ is integrable depends on $\alpha$. If $x$ is integrable, then $y$ will remain nonnegative if $y(0)$ is large enough.

The first equation is solvable in closed form, and then the second equation becomes a Riccati equation. For that, you have closed form solutions only for special values of $\alpha$. Some general observations: $y$ remains bounded if and only if it is always nonnegative. A necessary condition for that is that $y(0)$ is positive and $x$ is integrable. Whether $x$ is integrable depends on $\alpha$. If $x$ is not integrable, then $y$ cannot remain nonnegative. It may remain nonnegative if $x$ is integrable and $y(0)$ is large enough.

Source Link
Michael Renardy
  • 13k
  • 1
  • 42
  • 50

The first equation is solvable in closed form, and then the second equation becomes a Riccati equation. For that, you have closed form solutions only for special values of $\alpha$. Some general observations: $y$ remains bounded if and only if it is always nonnegative. A necessary condition for that is that $y(0)$ is positive and $x$ is integrable. Whether $x$ is integrable depends on $\alpha$. If $x$ is integrable, then $y$ will remain nonnegative if $y(0)$ is large enough.