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 1-30, solve the equation.
$t^{3} y^{2} dt + t^{4} y^{- 6} dy = 0$

$t = {Ce}^{(\frac{1}{7 y^{7}})}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition and concepts to be used

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation $F (x, y, \frac{dy}{dx}, ..., \frac{d^{n} y}{{dx}^{n}}) = 0$ we mean: Find a solution to the differential equation on an interval I that satisfies at x₀ the n initial conditions

$\begin{matrix} y (x_{0}) = y_{0}, \\ \frac{dy}{dx} (x_{0}) = y_{1}, \\ . \\ . \\ . \\ \frac{d^{n - 1} y}{{dx}^{n - 1}} (x_{0}) = y_{n - 1},, \end{matrix}$

Where $x_{0} \in I$ and $y_{0}, y_{1}, ..., y_{n - 1}$ are given constants.

Formulae to be used:

Integration by parts: $\int udv = uv - \int vdu$ .
$\int x^{a} dx = \frac{x^{a + 1}}{a + 1} + C$ .
$\int \frac{1}{x} dx = \ln |x| + C a$ .

Step 2: Given information and simplification

Given that, $t^{3} y^{2} dt + t^{4} y^{- 6} dy = 0 ...... (1)$

Evaluate the equation (1).

$\begin{matrix} t^{3} y^{2} dt + t^{4} y^{- 6} dy = 0 \\ t^{3} y^{2} dt = - t^{4} y^{- 6} dy \\ \frac{t^{3}}{t^{4}} dt = - \frac{y^{- 6}}{y^{2}} dy \\ \frac{1}{t} dt = - y^{- 8} dy ...... (2) \end{matrix}$

Now integrate the equation (2) on both sides.

$\int \frac{1}{t} dt = - \int y^{- 8} dy ...... (3)$

Step 3: Evaluation method

$\begin{matrix} \int \frac{1}{t} dt = - \int y^{- 8} dy \\ \ln |t| + C_{1} = \frac{1}{7} y^{- 7} \\ 7 \ln |t| + C = y^{- 7} \\ y = {(7 \ln |t| + C)}^{- \frac{1}{7}} \end{matrix}$

Take 7^th root on both sides.

$\begin{matrix} y^{7} = \frac{1}{7 \ln |t| + C} \\ y^{7} (7 \ln |t| + C) = 1 \\ t = {Ce}^{(\frac{1}{7 y^{7}})} \end{matrix}$

Hence, the solution of the given initial value problem is $t = {Ce}^{(\frac{1}{7 y^{7}})}$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 1-30, solve the equation.
$t^{3} y^{2} dt + t^{4} y^{- 6} dy = 0$

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

Step 3: Evaluation method

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 1-30, solve the equation. t3y2dt+t4y-6dy=0

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

Step 3: Evaluation method

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Question: In Problems 1-30, solve the equation.
$t^{3} y^{2} dt + t^{4} y^{- 6} dy = 0$