Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Question: In Problems 33–40, solve the equation given in: Problem 7.

The solution of the equation given in problem 7 is $c o s (x + y) + s i n (x + y) = C e^{x - y}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept and definition

When the right-hand side of the equation $\frac{d y}{d x} = f (x, y)$ can be expressed as a function of the combination $a x + b y$ , where a and b are constants, that is, $\frac{d y}{d x} = G (a x + b y)$ , then the substitution $z = a x + b y$ transforms the given equation into a separable equation.

Step2: Analyzing the given statement

One has to solve the equation given in problem 7, i.e.,

$\cos (x + y) d y = \sin (x + y) d x \cdot \cdot \cdot \cdot \cdot \cdot (1)$

Step 3: Rewriting the equation (1) in the form dydx=G(ax+by)

Rewriting the equation (1) as,

$\frac{d y}{d x} = \frac{\sin (x + y)}{\cos (x + y)}$

i.e., $\frac{d y}{d x} = \tan (x + y) \cdot \cdot \cdot \cdot \cdot \cdot (2)$

Step 4: Substituting z = x + y in equation (2) to transform the equation in the variables z and x

Taking $z = x + y$

Therefore, $d z = d x + d y$

Dividing both sides by dx,

$\begin{matrix} \frac{d z}{d x} = 1 + \frac{d y}{d x} \\ \frac{d y}{d x} = \frac{d z}{d x} - 1 \cdot \cdot \cdot \cdot \cdot \cdot (3) \end{matrix}$

Substitute z = x + y in equation (2) and the value of dy/dx from equation (3),

Step 5: Find the solution of equation dzdx=1+tanz

Separate the variables in equation (4),

$\begin{matrix} \frac{d z}{d x} - 1 = \tan z \\ \frac{d z}{d x} = 1 + \tan z \cdot \cdot \cdot \cdot \cdot \cdot (4) \\ \frac{d z}{1 + \tan z} = d x \end{matrix}$

Integrating both sides,

$\int \frac{d z}{1 + \tan z} = \int d x \cdot \cdot \cdot \cdot \cdot \cdot (5)$

Put tan z = t

$\begin{matrix} \sec^{2} z d z = d t \\ (1 + \tan^{2} z) d z = d t \\ (1 + t^{2}) d z = d t \\ d z = \frac{d t}{(1 + t^{2})} \end{matrix}$

Thus, equation (5) becomes,

$\int d x = \int \frac{d t}{(1 + t) (1 + t^{2})} \cdot \cdot \cdot \cdot \cdot \cdot (6)$

The right hand side of the equation (6) can be solved by using partial fraction,

Therefore,

$\frac{1}{(1 + t) (1 + t^{2})} = \frac{A}{1 + t} + \frac{B t + C_{1}}{1 + t^{2}}$

Multiply both sides by $(1 + t) (1 + t^{2})$

$1 = A (1 + t^{2}) + B t (1 + t) + C_{1} (1 + t) \cdot \cdot \cdot \cdot \cdot \cdot (7)$

Put t = -1

$\begin{matrix} 1 = 2 A + 0 + 0 \\ A = \frac{1}{2} \end{matrix}$

Equating coefficients in equation (7) of:

t²) $A + B = 0 \cdot \cdot \cdot \cdot \cdot \cdot (8)$

t ) $B + C_{1} = 0 \cdot \cdot \cdot \cdot \cdot \cdot (9)$

From equations (8) and (9),

$\begin{matrix} \frac{1}{2} + B = 0 \\ B = - \frac{1}{2} \\ - \frac{1}{2} + C_{1} = 0 \\ C_{1} = \frac{1}{2} \end{matrix}$

Therefore, equation (6) becomes,

$\begin{matrix} \int d x = \frac{1}{2} \int \frac{d t}{(1 + t)} - \frac{1}{2} \int \frac{t}{1 + t^{2}} d t + \frac{1}{2} \int \frac{d t}{1 + t^{2}} \\ \int d x = \frac{1}{2} \int \frac{d t}{(1 + t)} - \frac{1}{4} \int \frac{2 t}{1 + t^{2}} d t + \frac{1}{2} \int \frac{d t}{1 + t^{2}} \\ x = \frac{1}{2} \ln |1 + t| - \frac{1}{4} \ln |1 + t^{2}| + \frac{1}{2} \tan^{- 1} t + \ln C_{2} \end{matrix}$

Where, C₂ is the constant of integration.

Put t = tan z

$\begin{matrix} x = \frac{1}{2} \ln |1 + \tan z| - \frac{1}{4} \ln |1 + \tan^{2} z| + \frac{1}{2} \tan^{- 1} (\tan z) + \ln C_{2} \\ x = \frac{1}{2} \ln |1 + \tan z| - \frac{1}{4} \ln |1 + \tan^{2} z| + \frac{1}{2} z + \ln C_{2} \\ 2 x = \ln |1 + \tan z| - \frac{1}{2} \ln |1 + \tan^{2} z| + z + \ln C_{2} \\ 2 x - z = \ln |1 + \tan z| - \ln {(1 + \tan^{2} z)}^{1 / 2} + \ln C_{2} \end{matrix}$

$\begin{matrix} 2 x - z = \ln \frac{(1 + \tan z)}{{(1 + \tan^{2} z)}^{1 / 2}} + \ln C_{2} \\ 2 x - z = \ln \frac{C_{2} (1 + \tan z)}{{(1 + \tan^{2} z)}^{1 / 2}} \end{matrix}$

Put z = x + y

$\begin{matrix} 2 x - (x + y) = \ln \frac{C_{2} (1 + \tan (x + y))}{{(1 + \tan^{2} (x + y))}^{1 / 2}} \\ 2 x - x - y = \ln \frac{C_{2} (1 + \tan (x + y))}{{(1 + \tan^{2} (x + y))}^{1 / 2}} \\ x - y = \ln \frac{C_{2} (1 + \tan (x + y))}{{(1 + \tan^{2} (x + y))}^{1 / 2}} \\ e^{x - y} = \frac{C_{2} (1 + \frac{\sin (x + y)}{\cos (x + y)})}{{(1 + \frac{\sin^{2} (x + y)}{\cos^{2} (x + y)})}^{1 / 2}} \end{matrix}$

$\begin{matrix} e^{x - y} = \frac{C_{2} (\cos (x + y) + \sin (x + y))}{{(\cos^{2} (x + y) + \sin^{2} (x + y))}^{1 / 2}} \\ e^{x - y} = C_{2} (\cos (x + y) + \sin (x + y)) \\ \frac{e^{x - y}}{C_{2}} = (\cos (x + y) + \sin (x + y)) \\ \cos (x + y) + \sin (x + y) = C e^{x - y} \end{matrix}$

Where, $C = \frac{1}{C_{2}}$ , is an arbitrary constant.

Hence, the solution of the given equation $c o s (x + y) d y = s i n (x + y) d x$ is $c o s (x + y) + s i n (x + y) = C e^{x - y}$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 33–40, solve the equation given in: Problem 7.

Step by Step Solution

Step 1: Concept and definition

Step2: Analyzing the given statement

Step 3: Rewriting the equation (1) in the form dydx=G(ax+by)

Step 4: Substituting z = x + y in equation (2) to transform the equation in the variables z and x

Step 5: Find the solution of equation dzdx=1+tanz

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 33–40, solve the equation given in: Problem 7.

Step by Step Solution

Step 1: Concept and definition

Step2: Analyzing the given statement

Step 3: Rewriting the equation (1) in the form dydx=G(ax+by)

Step 4: Substituting z = x + y in equation (2) to transform the equation in the variables z and x

Step 5: Find the solution of equation dzdx=1+tanz

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