Chapter 2: Q7E (page 64)

In problems 1-8 Classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification.
$[2 x + ycos (xy)] dx + [xcos (xy) - 2 y] dy = 0$

Short Answer

Expert verified

The equation is exact.

Step by step solution

Find if the equation is separable

Here

$[2 x + y \cos (xy)] dx + [x \cos (xy) - 2 y] dy = 0 \frac{dy}{dx} = \frac{2 x + y \cos (xy)}{2 y - x \cos (xy)}$

This equation can be written in the form as g(x) h(y). so, this equation is not Separable

Determine if the equation is linear.

This equation is not linear of y and x because the RHS is not linear with respect to y and x respectively.

Evaluate whether the equation is exact.

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

$M (x, y) = 2 x + y \cos xy N (x, y) = x \cos xy - 2 y \frac{\partial M}{\partial y} = \cos xy - xy \sin xy = \frac{\partial N}{\partial x}$

This equation is exact.

Hence, the equation is exact.

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In problems 1-8 Classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification.
$[2 x + ycos (xy)] dx + [xcos (xy) - 2 y] dy = 0$

Short Answer

Step by step solution

Find if the equation is separable

Determine if the equation is linear.

Evaluate whether the equation is exact.

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In problems 1-8 Classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification.2x+ycos(xy)dx+xcos(xy)-2ydy=0

Short Answer

Step by step solution

Find if the equation is separable

Determine if the equation is linear.

Evaluate whether the equation is exact.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Statistics

Calculus

Logic and Functions

Mechanics Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In problems 1-8 Classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification.
$[2 x + ycos (xy)] dx + [xcos (xy) - 2 y] dy = 0$