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Chapter 6: Q. 36 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29鈥�52.
36. $\frac{d y}{d x} = \frac{6}{x}, y (1) = 7$

Short Answer

Expert verified

The answer is $y (x) = 6 \ln |x| + 7$

Step by step solution

01

Step 1. Given information

We have been given $\frac{d y}{d x} = \frac{6}{x}, y (1) = 7$

02

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

The differential equation can be solved by antidifferentiating.

$鈭� d y = 鈭� \frac{6}{x} d x y = 6 \ln |x| + C$

Now put $y (1) = 7$ ,

$7 = 6 \ln 1 + C 7 = C$

So, the answer is $y (x) = 6 \ln |x| + 7$

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Most popular questions from this chapter

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

$蟺 {鈭�}_{0}^{3} x^{2} d x a n d 蟺 {鈭�}_{0}^{3} (9 - x^{2}) d x$

$\frac{d y}{d x} = 3 - 4 y, y (0) = 2$

find an equation that gives y as an implicit function of x. Then draw the continuous curve that satisfies this differential equation and passes through the point (2, 0).

$\frac{d y}{d x} = \frac{x}{y}$

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

$\frac{d y}{d x} = 0.4 y - 100, y (0) = 0$

$\frac{d y}{d x} = 5 y, y (0) = 1$

See all solutions

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