Chapter 6: Q. 32 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
32. $\frac{d y}{d x} = - 5 y, y (0) = 2$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

We have been given $\frac{d y}{d x} = - 5 y, y (0) = 2$

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

The differential equation can be solved by antidifferentiating.

$\int \frac{1}{y} d y = \int - 5 d x \ln |y| = - 5 x + C y = e^{- 5 x + C} y = A e^{- 5 x} (e^{C} = A)$

Now put $y (0) = 2$ ,

$2 = A e^{- 5 (0)} 2 = A$

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

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.
32. $\frac{d y}{d x} = - 5 y, y (0) = 2$

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 29–52.32.dydx=-5y,y0=2

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.
32. $\frac{d y}{d x} = - 5 y, y (0) = 2$