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Chapter 2: Q4E (page 64)

In problems 1-8Classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification.

(yexy+2x)dx+(xexy-2y)dy=0

Short Answer

Expert verified

The equation is exact.

Step by step solution

01

Find if the equation is separable

Here

(yexy+2x)dx+(xexy-2y)dy=0dydx=(yexy+2x)(xexy-2y)

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

02

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.

03

Evaluate whether the equation is exact

The condition for exact is 鈭�M鈭�y=鈭�N鈭�x.

M(x,y)=(yexy+2x)N(x,y)=(xexy-2y)鈭�M鈭�y=exy+xyexy=鈭�N鈭�x

This equation is exact.

Hence, the equation is exact.

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Most popular questions from this chapter

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

dydx+xy=0

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16.(x2+y2)dx+2xydy=0

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9 -16

鈥�y2-xy鈥�dx+x2dy=0

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16鈥�xy+y2鈥�dx-x2dy=0

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

鈥�dydx+y鈥�tan鈥�x+sin鈥�x=0
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