Chapter 2: Q22E (page 64)

In Problems 21–26, solve the initial value problem.
$({ye}^{xy} - \frac{1}{y}) dx + ({xe}^{xy} + \frac{x}{y^{2}}) dy = 0, y (1) = 1$

Short Answer

Expert verified

$e^{xy} - \frac{x}{y} = e - 1$

Step by step solution

Evaluate the equation is exact

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

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

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

Therefore this equation is exact.

Find the value of F (x,y)

Here

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

Determine the value of g(y)

$\frac{\partial F}{\partial y} (x, y) = N (x, y) ({xe}^{xy} + \frac{x}{y^{2}}) + g' (y) = ({xe}^{xy} + \frac{x}{y^{2}}) g' (y) = 0 g (y) = C_{1}$

Now $F (x, y) = e^{xy} - \frac{x}{y} + C_{1}$

Therefore the solution of the differential equation is $e^{xy} - \frac{x}{y} = C$

Apply the initial conditions. $y (1) = 1$

$e^{1} - \frac{1}{1} = C C = e - 1$

Therefore, the solutions is role="math" localid="1664180725870" $e^{xy} - \frac{x}{y} = e - 1$ .

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In Problems 21–26, solve the initial value problem.
$({ye}^{xy} - \frac{1}{y}) dx + ({xe}^{xy} + \frac{x}{y^{2}}) dy = 0, y (1) = 1$

Short Answer

Step by step solution

Evaluate the equation is exact

Find the value of F (x,y)

Determine the value of g(y)

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In Problems 21–26, solve the initial value problem.yexy-1ydx+xexy+xy2dy=0,y(1)=1

Short Answer

Step by step solution

Evaluate 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

Calculus

Applied Mathematics

Discrete Mathematics

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Mechanics Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

In Problems 21–26, solve the initial value problem.
$({ye}^{xy} - \frac{1}{y}) dx + ({xe}^{xy} + \frac{x}{y^{2}}) dy = 0, y (1) = 1$