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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: Q12E (page 1)

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

Short Answer

Expert verified

The solution is $sin x cos {y +x}^{2} {-y}^{2} = C$

Step by step solution

01

Evaluate whether the equation is exact

Here $(\cos x \cos y + 2 x) dx - (\sin x \sin y + 2 y) dy = 0$

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

$M (x, y) = (\cos x \cos y + 2 x) N (x, y) = - (\sin x \sin y + 2 y)$

$\frac{鈭� M}{鈭� y} = - \cos x \sin y = \frac{鈭� N}{鈭� x}$

This equation is exact.

02

Find the value of F (x, y)

Here

$M (x, y) = (\cos x \cos y + 2 x) F (x, y) = 鈭� M (x, y) dx + g (y) = 鈭� (\cos x \cos y + 2 x) dx + g (y) = \sin x \cos y + x^{2} + g (y)$

03

determine the value of g(y)

$\frac{鈭� F}{鈭� y} (x, y) = N (x, y) - \sin x \sin y + g' (y) = - (\sin x \sin y + 2 y) g' (y) = - 2 y g (y) = - y^{2} + C_{1}$

Now $F(x,y) = sin x cos {y +x}^{2} {-y}^{2} {=C}_{1}$

The solution of the differential equation is $sin x cos {y +x}^{2} {-y}^{2} = C$

Hence the solution is $sin x cos {y +x}^{2} {-y}^{2} = C$ $sin x cos {y +x}^{2} {-y}^{2} = C$

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Most popular questions from this chapter

Verify that $蠒 (x) = \frac{2}{(1 - {ce}^{x})},$ where c is an arbitrary constant, it is a one-parameter family of solutions to $\frac{dy}{dx} = \frac{y (y - 2)}{2}$ . Graph the solution curves corresponding to $c = 0, 卤 1, 卤 2$ using the same coordinate axes.

Show that the equation ${(\frac{dy}{dx})}^{2} + y^{2} + 4 = 0$ has no (real-valued) solution.

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

Show that $蠒 (x) = {Ce}^{3 x} + 1$ is a solution tolocalid="1663944867164" $\frac{d y}{d x} - 3 y = - 3$ for any choice of the constant C. Thus, ${Ce}^{3 x} + 1$ is a one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes.

In Problem 19, solve the given initial value problem

$\begin{array}{l} y''' 鈭� y'' 鈭� 4 y' + 4 y = 0 \\ y (0) = 鈭� 4 \\ y' (0) = 鈭� 1 \\ y'' (0) = 鈭� 19 \end{array}$

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