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Q8E

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 7–9, solve the related phase plane differential equation (2), page 263, for the given system.**

**$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{{\mathit{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{2}}{{\mathit{y}}}^{\mathbf{-}\mathbf{3}}{\mathbf{,}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbf{3}}{{\mathit{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{2}}{\mathit{x}}{\mathit{y}}$**

**The solution is ${{\mathit{x}}}^{{\mathbf{3}}}{\mathbf{-}}{{\mathit{x}}}^{{\mathbf{2}}}{\mathit{y}}{\mathbf{-}}{{\mathit{y}}}^{\mathbf{-}\mathbf{2}}{\mathbf{=}}{\mathit{c}}$ .**

Here the system is;

$\frac{dx}{dt}={x}^{2}-2{y}^{-3}\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=3{x}^{2}-2xy$

And the phase plane equation is;

$\frac{dy}{dx}=\frac{3{x}^{2}-2xy}{{x}^{2}-2{y}^{-3}}$

Here the equation is $\frac{dy}{dx}=\frac{3{x}^{2}-2xy}{{x}^{2}-2{y}^{-3}}$.

$\begin{array}{rcl}(2xy-3{x}^{2})dx+({x}^{2}-2{y}^{-3})dy& =& 0\\ M& =& (2xy-3{x}^{2})\\ N& =& ({x}^{2}-2{y}^{-3})\\ \frac{\partial M}{\partial y}& =& 2x=\frac{\partial N}{\partial x}\end{array}$

** **Now,

$\begin{array}{rcl}F(x,y)& =& \int M(x,y)dx+g\left(y\right)\\ & =& \int (2xy-3{x}^{2})dx+g\left(y\right)\\ & =& {x}^{2}y-{x}^{3}+g\left(y\right)\\ N(x,y)& =& {x}^{2}+g\text{'}\left(y\right)\\ {x}^{2}-2{y}^{-3}& =& {x}^{2}+g\text{'}\left(y\right)\\ g\text{'}\left(y\right)& =& -2{y}^{-3}\\ g\left(y\right)& =& {y}^{-2}+c\\ F(x,y)& =& {x}^{3}-{x}^{2}y-{y}^{-2}+c\\ & & \end{array}$

Therefore, the solution is ${{\mathit{x}}}^{{\mathbf{3}}}{\mathbf{-}}{{\mathit{x}}}^{{\mathbf{2}}}{\mathit{y}}{\mathbf{-}}{{\mathit{y}}}^{\mathbf{-}\mathbf{2}}{\mathbf{=}}{\mathit{c}}$**.**

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