Chapter 4: Q4.3-14E (page 172)

Find a general solution. $y'' + 2 y' + 5 y = 0$

Short Answer

Expert verified

The general solution of the given equation $y'' + 2 y' + 5 y = 0$ is $y (t) = e^{- t} (c_{1} \cos (2 t) + c_{2} \sin (2 t))$ .

Step by step solution

Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $α \pm i β$ , then the general solution is given as:

$y (t) = c_{1} e^{α t} c o s β t + c_{2} e^{α t} s i n β t$ .

Finding roots of the auxiliary equation.

Given differential equation is $y'' + 2 y' + 5 y = 0 .$

Then the auxiliary equation is $r^{2} + 2 r + 5 = 0 .$

The roots of the auxiliary equation are:

role="math" localid="1654070409803" $r = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times 5}}{2 \times 1} r = \frac{- 2 \pm \sqrt{4 - 20}}{2} r = \frac{- 2 \pm \sqrt{- 16}}{2} r = \frac{- 2 \pm 4 i}{2} r = - 1 \pm 2 i$

Final answer.

Therefore, the general solution is:

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

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Find a general solution. $y'' + 2 y' + 5 y = 0$

Short Answer

Step by step solution

Complex conjugate roots.

Finding roots of the auxiliary equation.

Final answer.

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Find a general solution. y''+2y'+5y=0

Short Answer

Step by step solution

Complex conjugate roots.

Finding roots of the auxiliary equation.

Final answer.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Theoretical and Mathematical Physics

Calculus

Decision Maths

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

Find a general solution. $y'' + 2 y' + 5 y = 0$