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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: Q 3RP (page 79)

Question: In Problems 1-30, solve the equation.
$鈥� (x^{2} - 2 y^{- 3}) dy + (2 xy - 3 x^{2}) dx = 0$

Short Answer

Expert verified

The solution of the given equation is $x^{2} y - x^{3} + y^{- 2} = C$ .

Step by step solution

01

Given information and simplification

Given that, $(x^{2} - 2 y^{- 3}) dy + (2 xy - 3 x^{2}) dx = 0 鈰� 鈰� 鈰� 鈰� 鈰� 鈰� (1)$

Let us check whether the given equation is exact or not.

Then, $M = 2 xy - 3 x^{2}, N = x^{2} - 2 y^{- 3}$ .

Differentiate the value of M and N.

$\begin{array}{l} \frac{鈭� M}{鈭� y} = 2 x \\ \frac{鈭� N}{鈭� x} = 2 x \\ \frac{鈭� M}{鈭� y} = \frac{鈭� N}{鈭� x} \end{array}$

So, the given equation is exact.

02

Evaluation method

Now, let us assume $M = \frac{鈭� F}{鈭� x} = 2 xy - 3 x^{2}$ .

Integrate on both sides.

$\begin{matrix} F = 鈭� (2 xy - 3 x^{2}) dx \\ = x^{2} y - x^{3} + g (y) \end{matrix}$

Differentiate the F with respect to y.

$\begin{matrix} \frac{鈭� F}{鈭� y} = x^{2} + g' (y) \\ = N \end{matrix}$

Equalise the values of N.

$\begin{matrix} x^{2} + g' (y) = x^{2} - 2 y^{- 3} \\ g' (y) = - 2 y^{- 3} \end{matrix}$

Integrate on both sides.

$\begin{matrix} 鈭� g' (y) = - 鈭� 2 y^{- 3} dy \\ g (y) = y^{- 2} + C_{1} \end{matrix}$

Substitute in equation of F.

$\begin{matrix} x^{2} y - x^{3} + y^{- 2} + C_{1} = 0 \\ x^{2} y - x^{3} + y^{- 2} = C \end{matrix}$

So, the solution is role="math" localid="1664022924236" $x^{2} y - x^{3} + y^{- 2} = C$

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Most popular questions from this chapter

In problems, 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ . $\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}$

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

$鈥� t^{3} y^{2} dt + t^{4} y^{- 6} dy = 0$

Question: Magnetic Field Lines. As described in Problem 20 of Exercises 1.3, the magnetic field lines of a dipole satisfy.

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

Solve this equation and sketch several of these lines.

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

$\frac{dy}{dx} + xy = 0$

In problem 7-16, solve the equation. $\frac{dy}{dx} = \frac{\sec^{2} y}{1 + x^{2}}$

See all solutions

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