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Chapter 6: Q. 22 (page 575)

Solve the initial value problem
$\frac{d y}{d x} = y^{2} \sin (x), y (π) = 1$

Short Answer

Expert verified

The solution is $y (x) = \frac{1}{2 + \cos (x)}$

Step by step solution

01

Step 1. Given information

Given derivation is $\frac{d y}{d x} = y^{2} \sin (x), y (π) = 1$

02

Step 2. Solve the derivation

$\int \frac{1}{y^{2}} d y = \int \sin (x) d x - \frac{1}{y} = - \cos (x) + c y = \frac{1}{\cos (x) - c} y (π) = 1 1 = \frac{1}{- 1 - c} - 1 - c = 1 c = - 2 s u b s t i t u t e t h e c v a l u e i n e q u a t i o n, t h e s o l u t i o n i s y (x) = \frac{1}{2 + \cos (x)}$

03

Step 3. The solution

The solution is $y (x) = \frac{1}{2 + \cos (x)}$

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Most popular questions from this chapter

Find the exact value of the arc length of each function f (x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

$f (x) = 3 x + 1, [a, b] = [- 1, 4]$

Consider the region between the graph of $f (x) = \frac{3}{x}$ and the x-axis on [1,3]. For each line of rotation given in Exercises 31– 34, use definite integrals to find the volume of the resulting solid.

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.

34. $\frac{d y}{d x} = \frac{1}{y}, y (0) = 3$

For each pair of definite integrals in exercise 13-18 decide which if either is larger without computing the integrals

$π \int_{0}^{\frac{π}{2}} \cos^{2} (x) d x a n d π \int_{\frac{π}{2}}^{π} \cos^{2} (x) d x$

Consider the region between the graph of $f (x) = x - 2$ and the x-axis on [2,5]. For each line of rotation given in Exercises 35– 40, use definite integrals to find the volume of the resulting solid.

See all solutions

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