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

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $z'' - 6 z' + 10 z = 0$

Short Answer

Expert verified

The auxiliary equation for the given differential equation $z'' - 6 z' + 10 z = 0$ has complex roots and its general solution is $y (t) = e^{t} (c_{1} \cos 3 t + c_{2} \sin 3 t)$ .

Step by step solution

Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $α \pm iβ$ , 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 the roots of the auxiliary equation.

The differential equation is $z'' - 6 z' + 10 z = 0 .$

Then the auxiliary equation is $r^{2} - 6 r + 10 = 0 .$

The roots of the auxiliary equation are given below:

$r = \frac{6 \pm \sqrt{6^{2} - 4 \times 1 \times 10}}{2 \times 1} r = \frac{6 \pm \sqrt{36 - 40}}{} r = \frac{6 \pm 2 i}{} r = 3 \pm i$

Final answer.

Therefore, the general solution is:

$y (t) = e^{3 \times t} (c_{1} \cos (t) + c_{2} \sin (t)) = e^{t} (c_{1} \cos 3 t + c_{2} \sin 3 t)$

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The auxiliary equation for the given differential equation has complex roots. Find a general solution. $z'' - 6 z' + 10 z = 0$

Short Answer

Step by step solution

Complex conjugate roots.

Finding the roots of the auxiliary equation.

Final answer.

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The auxiliary equation for the given differential equation has complex roots. Find a general solution.z''-6z'+10z=0

Short Answer

Step by step solution

Complex conjugate roots.

Finding the roots of the auxiliary equation.

Final answer.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Probability and Statistics

Discrete Mathematics

Applied Mathematics

Decision Maths

Study anywhere. Anytime. Across all devices.

The auxiliary equation for the given differential equation has complex roots. Find a general solution. $z'' - 6 z' + 10 z = 0$