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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 5: Q8E (page 271)

In Problems 7鈥�9, solve the related phase plane differential equation (2), page 263, for the given system.
$\frac{d x}{d t} = x^{2} - 2 y^{- 3}, \frac{d y}{d t} = 3 x^{2} - 2 x y$

Short Answer

Expert verified

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

Step by step solution

01

Find phase plane equation

Here the system is;

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

And the phase plane equation is;

$\frac{d y}{d x} = \frac{3 x^{2} - 2 x y}{x^{2} - 2 y^{- 3}}$

02

Solve for exactness

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

$\begin{array}{rcl} (2 x y - 3 x^{2}) d x + (x^{2} - 2 y^{- 3}) d y & = & 0 \\ M & = & (2 x y - 3 x^{2}) \\ N & = & (x^{2} - 2 y^{- 3}) \\ \frac{鈭� M}{鈭� y} & = & 2 x = \frac{鈭� N}{鈭� x} \end{array}$

03

Find the value of F and G.

Now,

$\begin{array}{rcl} F (x, y) & = & 鈭� M (x, y) d x + g (y) \\ = & 鈭� (2 x y - 3 x^{2}) d x + g (y) \\ = & x^{2} y - x^{3} + g (y) \\ N (x, y) & = & x^{2} + g' (y) \\ x^{2} - 2 y^{- 3} & = & x^{2} + g' (y) \\ g' (y) & = & - 2 y^{- 3} \\ g (y) & = & y^{- 2} + c \\ F (x, y) & = & x^{3} - x^{2} y - y^{- 2} + c \end{array}$

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

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Most popular questions from this chapter

In Problems 3 鈥� 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$\frac{dx}{dt} + y = t^{2}, - x + \frac{dy}{dt} = 1$

In Problems 23 and 24, show that the given linear system is degenerate. In attempting to solve the system, determine whether it has no solutions or infinitely many solutions.

$D [x] + (D + 1) [y] = e^{t}, D^{2} [x] + (D^{2} + D) [y] = 0$

In Problems 1鈥�7, convert the given initial value problem into an initial value problem for a system in normal form.

$y^{6} (t) = {[y' (t)]}^{3} - \sin (y (t)) + e^{2 t}; y (0) = y' (0) = . . . . . . = y^{(5)} (0) = 0$

Solve the given initial value problem.

$x' = z - y; x (0) = 0 y' = z; 鈥� y (0) = 0 z' = z - x; z (0) = 2$

In Problems 11鈥�14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

$\frac{dx}{dt} = (y - x) (y - 1) \frac{dy}{dt} = (x - y) (x - 1)$

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