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Chapter 6: Q. 35 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29鈥�52.
35. $\frac{d y}{d x} = \frac{3}{y}, y (0) = 2$

Short Answer

Expert verified

The answer is $y (x) = \sqrt{6 x + 4}$

Step by step solution

01

Step 1. Given information

We have been given $\frac{d y}{d x} = \frac{3}{y}, y (0) = 2$

02

Step 2. Solve using antidifferentiation and/or variable separable method.

The differential equation can be solved by antidifferentiating.

$鈭� y d y = 鈭� 3 d x \frac{1}{2} y^{2} = 3 x + C_{1} y^{2} = 6 x + 2 C_{1} y^{2} = 6 x + C$

Now put $y (0) = 2$ ,

role="math" localid="1649147751077" ${(2)}^{2} = 6 (0) + C 4 = C$

So,

role="math" localid="1649147767750" $y^{2} = 6 x + 4 y = \sqrt{6 x + 4}$

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Most popular questions from this chapter

In the process of solving $\frac{d y}{d x} = y$ by separation of variables, we obtain the equation $| y | = e^{x + C}$ . After solving for $y$ , this equation becomes $y = A e^{x}$ . How is $A$ related to $C$ ? What happened to the absolute value?

Each of the definite integrals in Exercises 19鈥�24 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis. Find this region.

$蟺 {鈭�}_{0}^{2} ydy$

Each of the definite integrals in Exercises 19鈥�24 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis. Find this region.

$蟺 {鈭�}_{1}^{5} {(\frac{y - 1}{2})}^{2} dy$

Each of the definite integrals in Exercises 19鈥�24 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis. Find this region.

$蟺 {鈭�}_{0}^{4} (2^{2} - {(\sqrt{y})}^{2}) dy$

Each of the definite integrals in Exercises 19鈥�24 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis. Find this region.

$蟺 {鈭�}_{0}^{蟺} \sin^{2} xdx$

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