For x'=-x,\ y'=-2y, what happens as t→∞ to the solution starting at (1,1)?

It tends to the origin along the curve y=x^2

Exercise · ZW-PHASE-PLANE-007 · ★★☆ Intermediate

For \(x'=-x,\ y'=-2y\), what happens as \(t\to\infty\) to the solution starting at \((1,1)\)?

Choose one option.

Show answer

D. It tends to the origin along the curve \(y=x^2\)

Explanation

The solution is \(x=e^{-t},\ y=e^{-2t}=x^2\). Both coordinates tend to zero, so the trajectory enters the origin along \(y=x^2\).

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Review the concept: Phase plane

Coordinate plane \((x,y)\) of an autonomous system whose velocities \(f\) and \(g\) do not depend on \(t\); each solution is drawn there as a curve, its trajectory.

\[x'=f(x,y),\quad y'=g(x,y)\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{g(x,y)}{f(x,y)}\quad(f\neq0)\]

For \(x'=-x,\ y'=-2y\), what happens as \(t\to\infty\) to the solution starting at \((1,1)\)?

Explanation

Review the concept: Phase plane

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