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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: Q34E (page 77)

$x 鈮� 0$ In problems 33-40, Solve the equation given in Problem 2.
${(y - 4 x - 1)}^{2} dx - dy = 0$

Short Answer

Expert verified

$\frac{1}{4} \log |\frac{y - 4 x - 3}{y - 4 x + 1}| - x = C$ and

Step by step solution

01

Solving the given equation

The given equation is,

${(y - 4 x - 1)}^{2} dx - dy = 0$

Simplify the above equation,

$\begin{array}{l} {(y - 4 x - 1)}^{2} dx = dy \\ \frac{dy}{dx} = {(y - 4 x - 1)}^{2} 路路路路路路 (1) \end{array}$

Let,

$\begin{array}{l} t = y - 4 x - 1 \\ \frac{dt}{dx} = \frac{dy}{dx} - 4 \\ \frac{dy}{dx} = \frac{dt}{dx} + 4 \end{array}$

Substitute the $t = y - 4 x - 1$ and $\frac{dy}{dx} = \frac{dt}{dx} + 4$ in the equation (1),

$\frac{dt}{dx} + 4 = {(t)}^{2}$

Simplify the above equation,

$\frac{dt}{dx} = t^{2} - 4$

Cross multiplication on both sides in the above equation,

$\frac{dt}{t^{2} - 4} = dx$

02

Integrating

Taking integration of both sides in the above equation

$鈭� \frac{dt}{t^{2} - 2^{2}} = 鈭� dx$

One knows that,

$鈭� \frac{dx}{x^{2} - a^{2}} = \frac{1}{2 a} \log |\frac{x - a}{x + a}| + c$

Solve the above equation,

$\begin{array}{l} \frac{1}{2 (2)} \log |\frac{t - 2}{t + 2}| = x + c \\ \frac{1}{4} \log |\frac{t - 2}{t + 2}| - x = c \end{array}$

03

Finding the solution

Substitute the $t = y - 4 x - 1$ in the above equation,

$\frac{1}{4} \log |\frac{(y - 4 x - 1) - 2}{(y - 4 x - 1) + 2}| - x = c$

Simplify the above equation,

$\frac{1}{4} \log |\frac{y - 4 x - 3}{y - 4 x + 1}| - x = C$

Also, note that $x 鈮� 0$ is a solution.

Hence the solution is $\frac{1}{4} \log |\frac{y - 4 x - 3}{y - 4 x + 1}| - x = C$ and $x 鈮� 0$

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Most popular questions from this chapter

Question: In Problems 31-40, solve the initial value problem.

$(x^{3} - y) dx + xdy = 0, 鈥夆赌� y (1) = 3$

In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ . $(t + x + 2) dx + (3 t - x - 6) dt = 0$

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16. $\frac{dy}{诲胃} = \frac{胃 \sec \frac{y}{胃} + y}{胃}$

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

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

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

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