Chapter 4: Q13E (page 199)

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0. $13.9 t^{2} y'' (t) + 15 ty' (t) + y (t) = 0$

Short Answer

Expert verified

The general equation is $y = c_{1} t^{- 3} + c_{2} t^{- 3} \ln (t)$ .

Step by step solution

Find auxiliary equation.

The given differential equation is $9 t^{2} y'' (t) + 15 ty' (t) + y (t) = 0 \dots (1)$

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

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

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

$\begin{matrix} 9 t^{2} r (r - 1) t^{r - 2} + 15 {trt}^{r - 1} + t^{r} = 0 \\ 9 t^{2} r (r - 1) t^{r - 2} + 15 {trt}^{r - 1} + t^{r} = 0 \\ (9 r (r - 1) + 15 r + 1) t^{r} = 0 \\ 9 r^{2} + 6 r + 1 = 0 \end{matrix}$

Determine general equation.

The roots of the equation are:

$\begin{array}{l} r^{2} + 3 r + 3 r + 1 = 0 \\ (r + 3) (r + 3) = 0 \\ r = - 3, - 3 \end{array}$

The general equation is role="math" localid="1655207880130" $y = c_{1} t^{- 3} + c_{2} t^{- 3} \ln (t)$ .

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In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0. $13.9 t^{2} y'' (t) + 15 ty' (t) + y (t) = 0$

Short Answer

Step by step solution

Find auxiliary equation.

Determine general equation.

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In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.13.9t2y''(t)+15ty'(t)+y(t)=0

Short Answer

Step by step solution

Find auxiliary equation.

Determine general equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Calculus

Theoretical and Mathematical Physics

Decision Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0. $13.9 t^{2} y'' (t) + 15 ty' (t) + y (t) = 0$