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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 2: Q 33RP (page 79)

Question: In Problems 31-40, solve the initial value problem.
$(t + x + 3) dt + dx = 0, 鈥夆赌夆赌夆赌� x (0) = 1$

Short Answer

Expert verified

The solution of the given equation is $x = - t - 2 + 3 e^{- t}$ .

Step by step solution

01

Given information and simplification

Given that, $(t + x + 3) dt + dx = 0, 鈥夆赌夆赌夆赌� x (0) = 1 鈰� 鈰� 鈰� 鈰� 鈰� 鈰� (1)$

Evaluate the equation (1).

$\begin{matrix} (t + x + 3) dt + dx = 0 \\ dx = - (t + x + 3) dt \\ \frac{dx}{dt} = - (t + x + 3) \end{matrix}$

Let us assume $p = t + x + 3$ . Differentiate with respect to t.

$\frac{dp}{dt} = 1 + \frac{dx}{dt}$

Substitute the values in equation (1)

$\begin{matrix} \frac{dp}{dt} - 1 = - p \\ \frac{dp}{dt} = - p + 1 \\ \frac{1}{1 - p} dp = dt \end{matrix}$

Now integrate the equation (2) on both sides.

$\begin{matrix} 鈭� \frac{1}{1 - p} dp = 鈭� dt \\ \frac{\ln |1 - p|}{- 1} = t + C_{1} \\ \ln |1 - p| = - t + C_{1} \\ 1 - p = {Ce}^{- t} \end{matrix}$

02

Find the initial value

Substitute the value of p.

\begin{matrix} 1 - (t + x + 3) = {Ce}^{- t} \\ t + x + 2 = {Ce}^{- t} \end{matrix}

So, the solution is found.

Given that, x ( 0 ) = 1.

Then, x = 1 and t = 0.

Substitute the value in equation (4) to get the value of C.

$\begin{matrix} t + x + 2 = {Ce}^{- t} \\ 0 + 1 + 2 = {Ce}^{0} \\ 3 = C \end{matrix}$

Substitute the value of C in equation (2).

$\begin{matrix} t + x + 2 = 3 e^{- t} \\ x = - t - 2 + 3 e^{- t} \end{matrix}$

So, the solution is $x = - t - 2 + 3 e^{- t}$

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Most popular questions from this chapter

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16.

$\frac{dy}{dx} = \frac{y (lny - lnx + 1)}{x}$

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9- 16.

$鈥� (3 x^{2} - y^{2}) 鈥� dx + (xy - x^{3} y^{- 1}) dy = 0$

In problems identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y' = G (ax + by)$ . ${(y - 4 x - 1)}^{2} d x - d y = 0$

Question: In Problems 1 - 30, solve the equation.

$\frac{dy}{dx} - \frac{y}{x} = x^{2} \sin 鈥� 2 x$

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16. $\frac{dy}{dx} = \frac{x^{2} - y^{2}}{3 xy}$

See all solutions

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