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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 1: Q19E (page 1)

In Problems 9鈥�20, determine whether the equation is exact.
If it is, then solve it.
$(2 x + \frac{y}{1 + x^{2} y^{2}}) dx + (\frac{x}{1 + x^{2} y^{2}} - 2 y) dy = 0$

Short Answer

Expert verified

The solution is $x^{2} {-y}^{2} + arctan (xy) = C$ .

Step by step solution

01

Evaluate whether the equation is exact

Here $(2 x + \frac{y}{1 + x^{2} y^{2}}) dx + (\frac{x}{1 + x^{2} y^{2}} - 2 y) dy = 0$

The condition for exact is $\frac{鈭� M}{鈭� y} = \frac{鈭� N}{鈭� x}$ .

$M (x, y) = (2 x + \frac{y}{1 + x^{2} y^{2}}) N (x, y) = (\frac{x}{1 + x^{2} y^{2}} - 2 y) \frac{鈭� M}{鈭� y} = (\frac{1 - x^{2} y^{2}}{{(1 - x^{2} y^{2})}^{2}}) = \frac{鈭� N}{鈭� x}$

This equation is exact.

02

Find the value of F (x, y)

Here

$M (x, y) = (2 x + \frac{y}{1 + x^{2} y^{2}}) F (x, y) = 鈭� M (x, y) dx + g (y) = 鈭� (2 x + \frac{y}{1 + x^{2} y^{2}}) dx + g (y) = x^{2} + \tan^{- 1} (xy) + g (y)$

03

Determine the value of g(y)

Now $F (x, y) = x^{2} + \tan^{- 1} (xy) - y^{2} + C_{1}$

Therefore, the solution of the differential equation is

$x^{2} {+ tan}^{- 1} {(xy) -y}^{2} = C x^{2} {-y}^{2} + arctan (xy) = C$

Hence the solution is $x^{2} {-y}^{2} + arctan (xy) = C$

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Most popular questions from this chapter

In problems $1 - 4$ Use Euler鈥檚 method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

$\frac{dy}{dx} = y (2 - y)$ , $y (0) = 3$

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$6 - t^{3} (\frac{d^{2} v}{d t^{2}}) + 3 v - l n t = \frac{d v}{d t}$

Decide whether the statement made is True or False. The function $x (t) = t^{- 3} \sin t + 4 t^{- 3}$ is a solution to $t^{3} \frac{dx}{dt} = \cos t - 3 t^{2} x$ .

Implicit Function Theorem. Let $G (x, y)$ have continuous first partial derivatives in the rectangle $R = {(x, y) : a < x < b, c < y < d}$ containing the pointlocalid="1664009358887" $(x_{0}, y_{0})$ . If $G (x_{0}, y_{0}) = 0$ and the partial derivative $G_{y} (x_{0}, y_{0}) 鈮� 0$ , then there exists a differentiable function $y = 蠒 (x)$ , defined in some interval $I = (x_{0} - 未, y_{0} + 未)$ ,that satisfies G for allfor $G (x, 蠒 (x))$ all $x 鈭� I$ .

The implicit function theorem gives conditions under which the relationship $G (x, y) = 0$ implicitly defines yas a function of x. Use the implicit function theorem to show that the relationship $x + y + e^{xy} = 0$ given in Example 4, defines y implicitly as a function of x near the point $(0, - 1)$ .

In Problems $15 - 24$ , solve for $Y (s)$ , the Laplace transform of the solution $y (t)$ to the given initial value problem.

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