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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 20RP (page 79)

Question: In Problems 1-30, solve the equation.
$\frac{dy}{诲胃} + \frac{y}{胃} = - 4 {胃测}^{- 2}$

Short Answer

Expert verified

The solution of the given equation is $y = \sqrt[3]{- \frac{12 胃^{2}}{5} + \frac{C}{胃^{3}}}$ .

Step by step solution

01

Given information and simplification

Given that, $\frac{dy}{d 胃} + \frac{y}{胃} = - 4 胃 y^{- 2} 鈰� 鈰� 鈰� 鈰� 鈰� 鈰� (1)$

Multiply y² on both sides.

$y^{2} \frac{dy}{d 胃} + \frac{y^{3}}{胃} = - 4 胃$

Let us take v = y³. Then, $\frac{dv}{d 胃} = 3 y^{2} \frac{dy}{d 胃}$ .

$\frac{1}{3} \frac{dv}{d 胃} + \frac{v}{胃} = - 4 胃$

Multiply 3 and $胃^{3}$ on both sides.

$\begin{matrix} \frac{dv}{d 胃} 胃^{3} + 3 v 胃^{2} = - 12 胃^{4} \\ \frac{d}{d 胃} [v 胃^{3}] = - 12 胃^{4} \end{matrix}$

02

Evaluation method

Now integrate the equation (2) on both sides.

$\begin{matrix} 鈭� \frac{d}{d 胃} [v 胃^{3}] d 胃 = - 12 鈭� 胃^{4} d 胃 \\ v 胃^{3} = - \frac{12 胃^{5}}{5} + C \\ v = - \frac{12 胃^{2}}{5} + \frac{C}{胃^{3}} \end{matrix}$

Substitute the value of v.

$\begin{matrix} y^{3} = - \frac{12 胃^{2}}{5} + \frac{C}{胃^{3}} \\ y = \sqrt[3]{- \frac{12 胃^{2}}{5} + \frac{C}{胃^{3}}} \end{matrix}$

So, the solution is $y = \sqrt[3]{- \frac{12 胃^{2}}{5} + \frac{C}{胃^{3}}}$

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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. $(x^{2} + y^{2}) dx + 2 xy dy = 0$

Question: In Problems , solve the equation. $鈥� (4 {xy}^{3} - 9 y^{2} + 4 {xy}^{2}) dx + (3 x^{2} y^{2} - 6 xy + 2 x^{2} y) dy = 0$

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16 $鈥� (xy + y^{2}) 鈥� dx - x^{2} dy = 0$

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: 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 $b_{2} = k b_{1}$ .]

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