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Q5.4-1E

Expert-verifiedFound in: Page 271

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In Problems 1 and 2, verify that the pair x(t), and y(t) is a solution to the given system. Sketch the trajectory of the given solution in the phase plane.**

**$\frac{\mathbf{dx}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{3}}{{\mathbf{y}}}^{{\mathbf{3}}}{\mathbf{,}}\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{y}}{\mathbf{;}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{{\mathbf{e}}}^{\mathbf{3}\mathbf{t}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{{\mathbf{e}}}^{{\mathbf{t}}}$**

By putting the values of $\mathrm{x}\left(\mathrm{t}\right),\u200a\u200a\mathrm{y}\left(\mathrm{t}\right)$, get the result.

Here the system is

$\frac{dx}{dt}=3{y}^{3}\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=y$

And

$x\left(t\right)={e}^{3t}\phantom{\rule{0ex}{0ex}}y\left(t\right)={e}^{t}$

Then

role="math" localid="1663936766613" $\frac{dx}{dt}=3{e}^{3t}\phantom{\rule{0ex}{0ex}}=3{\left(y\left(t\right)\right)}^{3}\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}={e}^{t}\phantom{\rule{0ex}{0ex}}=y\left(t\right)$

Therefore, the pair is a solution to the system.

Also, ${\left(y\left(t\right)\right)}^{3}=x\left(t\right)={x}^{\frac{1}{3}}$

Since $\frac{dx}{dt}=3{y}^{3}$ **x is ****increasing ****when ****role="math" localid="1663936934373" ${\mathbf{y}}{\mathbf{>}}{\mathbf{0}}$**** ****and x is decreasing for**** role="math" localid="1663936952067" ${\mathit{y}}{\mathbf{<}}{\mathbf{0}}$**. This means the flow is from left to right along the part of the curve that lies above the x-axis, and the flow is from right to left along the part of the curve that lies below the x-axis.

This is the required result.

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