Chapter 4: Q4.3-10E (page 172)

Find a general solution $y'' + 4 y' + 8 y = 0$

Short Answer

Expert verified

The general solution of the given equation $y'' + 4 y' + 8 y = 0$ is $y (t) = e^{- 2 t} (c_{1} \cos (2 t) + c_{2} \sin (2 t))$ .

Step by step solution

Differentiate the value of y.

Given differential equation is $y'' + 4 y' + 8 y = 0 .$

Let role="math" localid="1654067583036" $y = e^{rt}$

Therefore, we have:

$y' (t) = r e^{rt}$

$y'' (t) = r^{2} e^{rt}$

Finding the roots.

Then the auxiliary equation is $r^{2} + 4 r + 8 = 0$

The roots of the auxiliary equation are:

$r = \frac{- 4 \pm \sqrt{4^{2} - 4 \times 1 \times 8}}{2 \times 1} r = \frac{- 4 \pm \sqrt{16 - 32}}{2} r = \frac{- 4 \pm 4 i}{2} r = - 2 \pm 2 i$

Final answer.

Therefore, the general solution is:

$y (t) = e^{- 2 \times t} (c_{1} \cos (2 t) + c_{2} \sin (2 t)) y (t) = e^{- 2 t} (c_{1} \cos (2 t) + c_{2} \sin (2 t))$

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Find a general solution $y'' + 4 y' + 8 y = 0$

Short Answer

Step by step solution

Differentiate the value of y.

Finding the roots.

Final answer.

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Find a general solution y''+4y'+8y=0

Short Answer

Step by step solution

Differentiate the value of y.

Finding the roots.

Final answer.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Calculus

Pure Maths

Geometry

Mechanics Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Find a general solution $y'' + 4 y' + 8 y = 0$