Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Solve the nonlinear differential equations in Exercises 6through 11 using the method of separation of variables:Write the differential equation $\frac{d x}{d t} = f x$ as $\frac{d x}{f x} = d t$ and integrate both sides.

8. $\frac{d x}{d t} = \sqrt{x}, x (0) = 4$

The solution is $x (t) = \frac{t^{2}}{4} + 4 + 2 t$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1 : Simplification for the differential equation

Consider the equation as follows.

$\frac{d x}{d t} = \sqrt{x}$

Now, separate the variables as follows.

$\begin{matrix} \frac{d x}{d t} = \sqrt{x} \\ \frac{d x}{\sqrt{x}} = d t \\ x^{\frac{- 1}{2}} d x = d t \end{matrix}$

Integrating on both sides as follows.

$\begin{matrix} x^{\frac{- 1}{2}} d x = d t \\ \int x^{\frac{- 1}{2}} d x = \int d t \\ \frac{x^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} = t + C \\ \frac{x^{\frac{- 1 + 2}{2}}}{\frac{- 1 + 2}{2}} = t + C \end{matrix}$

Simplify further as follows:

$\begin{matrix} \frac{x^{\frac{- 1 + 2}{2}}}{\frac{- 1 + 2}{2}} = t + C \\ \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = t + C \\ 2 x^{\frac{1}{2}} = t + C \end{matrix}$

Substituting the initial condition as follows.

$\begin{matrix} 2 x^{\frac{1}{2}} = t + C \\ 2 {(4)}^{\frac{1}{2}} = 0 + C {∵ x (0) = 4} \\ 2 (2) = C \\ 4 = C \end{matrix}$

Step 2 : Calculation of the solution

Now, substitute the value $4$ for $c$ in $2 x^{\frac{1}{2}} = t + C$ as follows:

$\begin{array}{l} 2 x^{\frac{1}{2}} = t + C \\ 2 x^{\frac{1}{2}} = t + 4 \\ x^{\frac{1}{2}} = \frac{1}{2} (t + 4) \\ x^{\frac{1}{2}} = \frac{t}{2} + \frac{4}{2} \end{array}$

Simplify further as follows.

$\begin{array}{l} x^{\frac{1}{2}} = \frac{t}{2} + \frac{4}{2} \\ x^{\frac{1}{2}} = \frac{t}{2} + 2 \end{array}$

Now, squaring on both sides as follows:

$\begin{matrix} x^{\frac{1}{2}} = \frac{t}{2} + 2 \\ {(x^{\frac{1}{2}})}^{2} = {(\frac{t}{2} + 2)}^{2} \\ x = \frac{t^{2}}{4} + 4 + 2 \times 2 \times \frac{t}{2} \\ x (t) = \frac{t^{2}}{4} + 4 + 2 t \end{matrix}$

Hence, the solution for the differential equation $\frac{d x}{d t} = \sqrt{x}$ is $x (t) = \frac{t^{2}}{4} + 4 + 2 t$ .

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Linear Algebra With Applications

Answers without the blur.

Short Answer

Solve the nonlinear differential equations in Exercises 6through 11 using the method of separation of variables:Write the differential equation $\frac{d x}{d t} = f x$ as $\frac{d x}{f x} = d t$ and integrate both sides.

8. $\frac{d x}{d t} = \sqrt{x}, x (0) = 4$

Step by Step Solution

Step 1 : Simplification for the differential equation

Step 2 : Calculation of the solution

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Linear Algebra With Applications

Answers without the blur.

Short Answer

Solve the nonlinear differential equations in Exercises 6through 11 using the method of separation of variables:Write the differential equation dxdt=fx as dxfx=dtand integrate both sides. 8.dxdt=x,x(0)=4

Step by Step Solution

Step 1 : Simplification for the differential equation

Step 2 : Calculation of the solution

Most popular questions for Math Textbooks

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Pure Maths

Solve the nonlinear differential equations in Exercises 6through 11 using the method of separation of variables:Write the differential equation $\frac{d x}{d t} = f x$ as $\frac{d x}{f x} = d t$ and integrate both sides.

8. $\frac{d x}{d t} = \sqrt{x}, x (0) = 4$