Chapter 4: Q11E (page 199)

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.
$t^{2} \frac{d^{2} z}{{dt}^{2}} + 5 t \frac{dz}{dt} + 4 z = 0$

Short Answer

Expert verified

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

Step by step solution

Find the auxiliary equation.

Given differential equation $t^{2} \frac{d^{2} z}{{dt}^{2}} + 5 t \frac{dz}{dt} + 4 z = 0 \dots (1)$

Assume $z = t^{r}$ then we have:

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

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

$\begin{matrix} t^{2} r (r - 1) t^{r - 2} + 5 {trt}^{r - 1} + 4 t^{r} = 0 \\ (r (r - 1) + 5 r + 4) t^{r} = 0 \\ r^{2} + 4 r + 4 = 0 \end{matrix}$

Determine the general equation.

The roots of the equation are:

$\begin{array}{l} r^{2} + 2 r + 2 r - 6 = 0 \\ (r + 2) (r + 2) = 0 \\ r = - 2, 2 \end{array}$

Thus, the general equation is $z = c_{1} t^{- 2} + c_{2} t^{- 2} lnt$ .

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In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.
$t^{2} \frac{d^{2} z}{{dt}^{2}} + 5 t \frac{dz}{dt} + 4 z = 0$

Short Answer

Step by step solution

Find the auxiliary equation.

Determine the general equation.

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In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.t2d2zdt2+5tdzdt+4z=0

Short Answer

Step by step solution

Find the auxiliary equation.

Determine the general equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Pure Maths

Discrete Mathematics

Geometry

Statistics

Study anywhere. Anytime. Across all devices.

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.
$t^{2} \frac{d^{2} z}{{dt}^{2}} + 5 t \frac{dz}{dt} + 4 z = 0$