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Expert-verified Found in: Page 271 ### Fundamentals Of Differential Equations And Boundary Value Problems

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

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

See the step by step solution

## Step 1: Find phase plane equation

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;

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

## Step 2: Solve for exactness

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

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

## Step 3: Find the value of F and G.

Now,

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

Therefore, the solution is ${{\mathbit{x}}}^{{\mathbf{3}}}{\mathbf{-}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbit{y}}{\mathbf{-}}{{\mathbit{y}}}^{\mathbf{-}\mathbf{2}}{\mathbf{=}}{\mathbit{c}}$. ### Want to see more solutions like these? 