Chapter 6: Q. 31 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
31. $\frac{d y}{d x} = - 6 x^{2}, y (1) = 3$

Short Answer

Expert verified

The answer is $y (x) = - 2 x^{3} + 5$

Step by step solution

Step 1. Given information

We have been given $\frac{d y}{d x} = - 6 x^{2}, y (1) = 3$

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

The differential equation can be solved by antidifferentiating.

$\int d y = = \int - 6 x^{2} d x y = - 6 \cdot \frac{1}{3} x^{3} + C y = - 2 x^{3} + C$

Now put $y (1) = 3$ ,

$3 = - 2 {(1)}^{3} + C 3 + 2 = C 5 = C$

So, the answer is $y (x) = - 2 x^{3} + 5$

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
31. $\frac{d y}{d x} = - 6 x^{2}, y (1) = 3$

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 29–52.31.dydx=-6x2,y1=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

Recommended explanations on Math Textbooks

Logic and Functions

Probability and Statistics

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Study anywhere. Anytime. Across all devices.

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
31. $\frac{d y}{d x} = - 6 x^{2}, y (1) = 3$