Chapter 1: Q19E (page 1)

In Problems 9–20, 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

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 $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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

This equation is exact.

Find the value of F (x, y)

Here

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

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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In Problems 9–20, 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

Step by step solution

Evaluate whether the equation is exact

Find the value of F (x, y)

Determine the value of g(y)

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In Problems 9–20, determine whether the equation is exact.If it is, then solve it.2x+y1+x2y2dx+x1+x2y2-2ydy=0

Short Answer

Step by step solution

Evaluate whether the equation is exact

Find the value of F (x, y)

Determine the value of g(y)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Applied Mathematics

Probability and Statistics

Decision Maths

Discrete Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

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