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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 2: Q21E (page 64)

In Problems 21鈥�26, solve the initial value problem.
$(\frac{1}{x} + 2 y^{2} x) dx + (2 y x^{2} - cosy) dy = 0, y (1) = 蟺$

Short Answer

Expert verified

The solution is $\ln |x| + x^{2} y^{2} - \sin y = 蟺^{2}$ .

Step by step solution

01

Evaluate the equation is exact

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

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

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

$\frac{鈭� M}{鈭� y} = 4 xy = \frac{鈭� N}{鈭� x}$

This equation is exact.

02

Find the value of F(x, y)

Here

M (x, y) = (\frac{1}{x} + 2 y^{2} x) F (x, y) = 鈭� M (x, y) dx + g (y) = 鈭� (\frac{1}{x} + 2 y^{2} x) dx + g (y) = \ln |x| + x^{2} y^{2} + g (y)

03

Determine the value of g(y)

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

Now $F (x, y) = \ln |x| + x^{2} y^{2} - \sin y$

The solution of the differential equation is $\ln |x| + x^{2} y^{2} - \sin y = C$ .

Apply the initial conditions.

$\ln |1| + 1^{2} 蟺^{2} - \sin 蟺 = C C = 蟺^{2}$

The solution is $\ln |x| + x^{2} y^{2} - \sin y = 蟺^{2}$ .

Hence the solution is role="math" localid="1664179448399" $\ln | x | + x^{2} y^{2} - \sin y = 蟺^{2}$

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Most popular questions from this chapter

Question: In Problems 1-30, solve the equation.

$(3 x - y - 5) dx + (x - y + 1) dy = 0$

In problem $1 - 6$ , determine whether the given differential equation is separable $s^{2} + \frac{d s}{d t} = \frac{s + 1}{s t}$ .

In problems 33-40, Solve the equation given in Problem 1.

$2 t x 鈥� d x + (t^{2} - x^{2}) d t = 0$

In problem $1 - 6,$ determine whether the differential equation is separable $\frac{d y}{d x} = 4 y^{2} - 3 y + 1$ .

Use the method discussed under 鈥淓quations of the Form $\frac{dy}{dx} = G (ax + by)$ " to solve problems 17-20.

$\frac{dy}{dx} = \sqrt{x + y} - 1$

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