Chapter 4: Q4.3-19E (page 172)

Find a general solution $y''' + y'' + 3 y' - 5 y = 0$

Short Answer

Expert verified

The general solution of the given equation $y''' + y'' + 3 y' - 5 y = 0$ is $y (t) = C_{1} e^{t} + C_{2} e^{- t} \cos (2 t) + C_{3} e^{- t} \sin (2 t)$ .

Step by step solution

Using rational root theorem.

First, one needs to find the auxiliary equation and solve it. One has $r^{3} + r^{2} + 3 r - 5 = 0$ .

You can use the rational root theorem.

The first divisor of role="math" localid="1654073691473" $5$ is role="math" localid="1654073695156" $1$ if will be one solution of the equation and $(r - 1)$ will be a factor.

Indeed, you have $1^{3} + 1^{2} + 3 - 5 = 0$

Now you can dividerole="math" localid="1654073800144" $r^{3} + r^{2} + 3 r - 5$ byrole="math" localid="1654073795414" $r - 1$ to get $r^{2} + 2 r + 5$ .

Finding factors and roots.

The equation can be factored as $(r - 1) (r^{2} + 2 r + 5) = 0 .$

Since $r^{2} + 2 r + 5 = 0$

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

The roots of the auxiliary equation are $r = 1, r = - 1 + 2 i$ and $r = - 1 - 2 i$

Thus, the general solution of the differential equation is: $y (t) = C_{1} e^{t} + C_{2} e^{- t} \cos (2 t) + C_{3} e^{- t} \sin (2 t)$

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Find a general solution $y''' + y'' + 3 y' - 5 y = 0$

Short Answer

Step by step solution

Using rational root theorem.

Finding factors and roots.

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Find a general solution y'''+y''+3y'-5y=0

Short Answer

Step by step solution

Using rational root theorem.

Finding factors and roots.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Find a general solution $y''' + y'' + 3 y' - 5 y = 0$