Chapter 6: Q. 34 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
34. $\frac{d y}{d x} = \frac{1}{y}, y (0) = 3$

Short Answer

Expert verified

The answer is $y (x) = \sqrt{2 x + 9}$

Step by step solution

Step 1. Given information

We have been given $\frac{d y}{d x} = \frac{1}{y}, y (0) = 3$

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

The differential equation can be solved by antidifferentiating.

$\int y d y = \int d x \frac{1}{2} y^{2} = x + C_{1} y^{2} = 2 x + C (2 C_{1} = C)$

Now put $y (0) = 3$ ,

${(3)}^{2} = 2 (0) + C 9 = C$

So,

$y^{2} = 2 x + 9 y = \sqrt{2 x + 9}$

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
34. $\frac{d y}{d x} = \frac{1}{y}, y (0) = 3$

Short Answer

Step by step solution

Step 1. Given information

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

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.34.dydx=1y,y0=3

Short Answer

Step by step solution

Step 1. Given information

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

One App. One Place for Learning.

Most popular questions from this chapter

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
34. $\frac{d y}{d x} = \frac{1}{y}, y (0) = 3$