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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 1: Q10RP (page 1)

Decide whether the statement made is True or False. The relation $\sin y + e^{y} = x^{6} - x^{2} + x + 1$ is an implicit solution to $\frac{dy}{dx} = \frac{6 x^{5} - 2 x + 1}{\cos y + e^{y}}$ .

Short Answer

Expert verified

The statement is true.

Step by step solution

01

Using the differential formula

For the result use the differential formula $\frac{d}{dx} (x^{n}) = n x^{n - 1}$ and consider x and y as variable.

02

Differentiating sin y+ey=x6-x2+x+1 with respect to x.

The Solution is given below,

$\frac{d}{dx} (\sin y + e^{y}) = \frac{d}{dx} (x^{6} - x^{2} + x + 1) cosy \frac{dy}{dx} + e^{y} \frac{dy}{dx} = 6 x^{5} - 2 x + 1 (cosy + e^{y}) \frac{dy}{dx} = 6 x^{5} - 2 x + 1 \frac{dy}{dx} = \frac{6 x^{5} - 2 x + 1}{cosy + e^{y}}$

Hence, this is the given differential equation, the given statement is true.

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Most popular questions from this chapter

Show that $蠒 (x) = c_{1} \sin x + c_{2} \cos x,$ is a solution to $\frac{d^{2} y}{{dx}^{2}} + y = 0$ for any choice of the constants $c_{1}$ and $c_{2}$ . Thus, $c_{1} \sin x + c_{2} \cos x,$ is a two-parameter family of solutions to the differential equation.

In Problems 21鈥�26, solve the initial value problem.

$(e^{t} y + {te}^{t} y) dt + (t e^{t} + 2) dy = 0, y (0) = - 1$

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$3 r - c o s 胃 \frac{d r}{d 胃} = s i n 胃$

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(a) 30 minutes.

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