Chapter 4: Q37E (page 165)

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it.
$y''' + y'' - 6 y' + 4 y = 0$ .

Short Answer

Expert verified

$y (t) = c_{1} e^{t} + c_{2} e^{(- 1 - \sqrt{5}) t} + c_{3} e^{(- 1 + \sqrt{5}) t}$ is the solution of the given equation $y''' + y'' - 6 y' + 4 y = 0$

Step by step solution

Differentiate the value of y

Given differential equation is $y''' + y'' - 6 y' + 4 y = 0$

Let, $y = e^{r t}$ then $y' (t) = r e^{r t}$ and $y (t) = r^{2} e^{r t}$

$y''' = r^{3} e^{r t}$

Then the auxiliary equation is $r^{3} e^{r t} + r^{2} e^{r t} - 6 r e^{r t} + 4 e^{r t} = 0$

$(r^{3} + r^{2} - 6 r + 4) e^{r t} = 0 r^{3} + r^{2} - 6 r + 4 = 0$

Substitute the value for r

Let substitute $r = 1$ then one gets $1 + 1 - 6 + 4 = 0$

Therefore $r = 1$ is the solution of the auxiliary equation $r^{3} + r^{2} - 6 r + 4 = (r - 1) (r^{2} + 2 r - 4)$

$(r - 1) (r^{2} + 2 r - 4) = 0$

Hence, the solutions are;

$r = 1$

And

$r = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times - 4}}{2 \times 1} r = \frac{- 2 \pm \sqrt{20}}{2 \times 1} r = - 1 \pm \sqrt{5}$

So, the solutions are $y (t) = c_{1} e^{t} + c_{2} e^{(- 1 - \sqrt{5}) t} + c_{3} e^{(- 1 + \sqrt{5}) t}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it.
$y''' + y'' - 6 y' + 4 y = 0$ .

Short Answer

Step by step solution

Differentiate the value of y

Substitute the value for r

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Applied Mathematics

Decision Maths

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it.y'''+y''-6y'+4y=0.

Short Answer

Step by step solution

Differentiate the value of y

Substitute the value for r

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Applied Mathematics

Decision Maths

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it.
$y''' + y'' - 6 y' + 4 y = 0$ .