Chapter 1: Q25E (page 1)

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

Short Answer

Expert verified

The solution is $\sin x - x \cos x = \ln |y| + \frac{1}{y} + π - 1$ .

Step by step solution

Step 1: Evaluate the equation is exact

Here $(y^{2} \sin x) dx + (\frac{1}{x} - \frac{y}{x}) dy, y (π) = 1$

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

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

This equation is not exact.

Find the solution

But this equation can be separable.

$(y^{2} \sin x) dx + (\frac{1 - y}{x}) dy = 0 (x \sin x) dx = (\frac{y - 1}{y^{2}}) dy$

Integrating on both sides, the result is

$- x \cos x + \sin x = \ln |y| + \frac{1}{y} + C$

Therefore, the solution of the differential equation is $- x \cos x + \sin x = \ln |y| + \frac{1}{y} + C$

Apply the initial conditions

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

$- π \cos π + \sin π = \ln |1| + \frac{1}{1} + C C = π - 1$

The solutions is

$- x \cos x + \sin x = \ln |y| + \frac{1}{y} + π - 1 \sin x - x \cos x = \ln |y| + \frac{1}{y} + π - 1$

Hence the solution is role="math" localid="1664190043647" $\sin x - x \cos x = \ln |y| + \frac{1}{y} + π - 1$

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

Short Answer

Step by step solution

Step 1: Evaluate the equation is exact

Find the solution

Apply the initial conditions

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In Problems 21–26, solve the initial value problem.(y2sinx)dx+(1x-yx)dy,y(π)=1

Short Answer

Step by step solution

Step 1: Evaluate the equation is exact

Find the solution

Apply the initial conditions

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Discrete Mathematics

Logic and Functions

Geometry

Theoretical and Mathematical Physics

Calculus

Study anywhere. Anytime. Across all devices.

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