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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: Q38RP (page 79)

In Problems , solve the initial value problem.
$\sqrt{y} dx + (x^{2} + 4) dy = 0, 鈥� 鈥� 鈥� 鈥� y (0) = 4$

Short Answer

Expert verified

The solution of the given equation is $y = {(- \frac{1}{4} \arctan (\frac{x}{2}) + 2)}^{2}$ .

Step by step solution

01

Given information and simplification

Given that, $\sqrt{y} dx + (x^{2} + 4) dy = 0, 鈥� 鈥� 鈥� 鈥� y (0) = 4 路路路路路路 (1)$

Evaluate the equation (1).

$\sqrt{y} dx + (x^{2} + 4) dy = 0 (x^{2} + 4) dy = - \sqrt{y} dx \frac{dy}{dx} = - \frac{\sqrt{y}}{x^{2} + 4} \frac{dy}{\sqrt{y}} = - \frac{dx}{x^{2} + 4}$

Now integrate the equation (2) on both sides.

$鈭� \frac{dy}{\sqrt{y}} = - 鈭� \frac{dx}{x^{2} + 4} 2 \sqrt{y} = - \frac{1}{2} \arctan (\frac{x}{2}) + C_{1} \sqrt{y} = - \frac{1}{4} \arctan (\frac{x}{2}) + C 路路路路路路 (2)$

02

Find the initial value

Given that, $y (0) = 4$ .

Then, x = 0 and y = 4.

Substitute the value in equation (2) to get the value of C.

$\sqrt{y} = - \frac{1}{4} \arctan (\frac{x}{2}) + C \sqrt{4} = - \frac{1}{4} \arctan (\frac{0}{2}) + C 2 = C$

Substitute the value of C in equation (2).

$\sqrt{y} = - \frac{1}{4} \arctan (\frac{x}{2}) + 2 y = {(- \frac{1}{4} \arctan (\frac{x}{2}) + 2)}^{2}$

So, the solution is, $y = {(- \frac{1}{4} \arctan (\frac{x}{2}) + 2)}^{2}$

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Most popular questions from this chapter

Question: In Problems , solve the equation.

$\frac{dy}{dx} = 2 - \sqrt{2 x - y + 3}$

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

$(y - 2 x - 1) dx + (x + y - 4) dy = 0$

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

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

Question: Show that equation (13) reduces to an equation of the form $\frac{d y}{d x} = G (a x + b y),$

when [Hint: If $a_{1} b_{2} = a_{2} b_{1}$ , then $a_{2} / a_{1} = b_{2} / b_{1} = k,$ so that $a_{2} = k a_{1}$ and $b_{2} = k b_{1}$ .]

In problem 7-16, solve the equation. $\frac{dx}{dt} = \frac{t}{x 鈥� e^{t + 2 x}}$

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