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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 4: Q12E (page 199)

In Problems 9 through 14, find a general solution to the given Cauchy鈥揈uler equation for t>0.
$12 . \frac{d^{2} w}{{dt}^{2}} + \frac{6}{t} \frac{dw}{dt} + \frac{4}{t^{2}} w = 0$

Short Answer

Expert verified

The general equation is $w = c_{1} t^{- 4} + c_{2} t^{- 1}$ .

Step by step solution

01

Find the auxiliary equation.

Given differential equation $\frac{d^{2} w}{{dt}^{2}} + \frac{6}{t} \frac{dw}{dt} + \frac{4}{t^{2}} w = 0 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� 鈥� (1)$

Assume $w = t^{r}$ then we have;

$\begin{array}{l} w' = {rt}^{r - 1} \\ w'' = r (r - 1) t^{r - 2} \end{array}$

Substitute all values in equation (1), and we get:

role="math" localid="1655206825754" $\begin{matrix} r (r - 1) t^{r - 2} + \frac{6}{t} {rt}^{r - 1} + \frac{4}{t^{2}} t^{r} = 0 \\ t^{2} r (r - 1) t^{r - 2} + 6 {trt}^{r - 1} + 4 t^{r} = 0 \\ (r (r - 1) + 6 r + 4) t^{r} = 0 \\ r^{2} + 5 r + 4 = 0 \end{matrix}$

02

Determine the general equation.

The roots of the equation are:

$\begin{array}{l} r^{2} + 4 r + r + 4 = 0 \\ (r + 4) (r + 1) = 0 \\ r = - 4, - 1 \end{array}$

The general equation is $w = c_{1} t^{- 4} + c_{2} t^{- 1}$ .

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Most popular questions from this chapter

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $y'' - 4 y' + 7 y = 0$

Find a general solution $u'' + 7 u = 0$

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $y'' - 10 y' + 26 y = 0$

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $4 y'' + 4 y' + 6 y = 0$

Solve the given initial value problem. $y''' - 4 y'' + 7 y' - 6 y = 0;$ $y (0) = 1, y' (0) = 0,$ $y'' (0) = 0$

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