Chapter 6: Q. 43 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52
$\frac{d y}{d x} = 9.8 - \frac{0.3}{150} y, y (0) = 1000$

Short Answer

Expert verified

The solution of the initial-value problem $\frac{d y}{d x} = 9.8 - \frac{0.3}{150} y, y (0) = 1000 a s y (x) = 4900 - 3900 e^{\frac{- x}{500}}$

Step by step solution

Step 1. Given information

The given initial value problem $\frac{d y}{d x} = 9.8 - \frac{0.3}{150} y, y (0) = 1000 . . . . . . . . . . . (1)$

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

Note that the differential equation in (1) does not contain the independent variable at all, so technically the variables have already been separated. Hence, the differential equation can be solved by antidifferentiating. Thus, the solution of the differential equation involved in the initialvalue problem is given by

\int \frac{d y}{9.8 - \frac{0.3}{150} y} = \int d x - \frac{150}{0.3} \ln | 9.8 - \frac{0.3}{150} y | = x + C_{1} \ln | 9.8 - \frac{0.3}{150} y | = - \frac{0.3}{150} x + C (- \frac{0.3}{150} C_{1} = C) 9.8 - \frac{0.3}{150} y = A e^{\frac{0.3}{150} x} (e^{c} = A)

Simplify the above expression further

\frac{0.3}{150} y = 9.8 - A e^{\frac{0.3}{150} x} y = 500 [9.8 - A e^{- \frac{1}{500} x}] = 4900 - 500 A e^{- \frac{1}{500} x}

Now, use the given initial condition

y (0) = 1000

, that is take

x = 0, y = 1000

in the above result and evaluate the constant A.

$1000 = 4900 - 500 A 500 A = 3900 A = \frac{39}{5}$

Substitute this value of the constant Ain the solution of the differential equation and obtain the solution of the initial-value problem $\frac{d y}{d x} = 9.8 - \frac{0.3}{150} y, y (0) = 1000 a s y (x) = 4900 - 3900 e^{\frac{- x}{500}}$

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52
$\frac{d y}{d x} = 9.8 - \frac{0.3}{150} y, y (0) = 1000$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52dydx=9.8-0.3150y,y(0)=1000

Short Answer

Step by step solution

Step 1. Given information

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Decision Maths

Theoretical and Mathematical Physics

Discrete Mathematics

Statistics

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
$\frac{d y}{d x} = 9.8 - \frac{0.3}{150} y, y (0) = 1000$