Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: Find phase plane equation

Here the system is;

$$\begin{gathered} \frac{dx}{dt}=x^2-2y^{-3} \\ \frac{dy}{dt}=3x^2-2xy \end{gathered}$$ 

And the phase plane equation is;

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

Step 2: Solve for exactness

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

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

Step 3: Find the value of F and G.

 Now, 

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

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