Mathematics College

Find the general solution of y''' + 9y'' + y' − 111y = 0 if it is known that y1 = e−6x cos(x) is one solution.

Answers

Answer 1

Answer:

[tex]\mathbf{y(x) = \dfrac{(c_1 + c_2) \ cos (x)}{e^{6x} }+ \dfrac{i(c_1 -c_2) \ sin (x)}{e^{6x}}+ c_3e^{3\ x}}[/tex]

Step-by-step explanation:

The objective of this question is to solve:

[tex]\dfrac{dy(x)}{dx}+ 9\dfrac{d^2y(dx)}{dx^2}+\dfrac{d^3y(x)}{dx^3}-111y(x) = 0 :[/tex]

Suppose the general solution is proportional to [tex]e^{\lambda x}[/tex] for [tex]\lambda[/tex] is constant; Then:

Let's replace [tex]y(x) = e^{\lambda\ x}[/tex] into the above equation:

i.e.

[tex]\dfrac{d^3}{dx^3}(e ^{\lambda x} )+ 9 \dfrac{d^2}{dx^2}(e ^{\lambda x} ) + \dfrac{d}{dx}(e ^{\lambda x} )- 111 \ e ^{\lambda x} = 0[/tex]

To Replace:

[tex]\dfrac{d^3}{dx^3}(e ^{\lambda x} )[/tex] with [tex]\lambda^3 e ^{\lambda x }[/tex]

[tex]\dfrac{d^2}{dx^2}(e ^{\lambda x} ) \ with \ \lambda^2 e^{\lambda\ x}[/tex]

[tex]\dfrac{d}{dx}(e ^{\lambda x} ) \ with \ \lambda e ^{\lambda \ x}[/tex]

Thus;

[tex]\lambda^3 e ^{\lambda x }[/tex] + [tex]9 \lambda^2 e^{\lambda\ x}[/tex] + [tex]\lambda e ^{\lambda \ x}[/tex] - 111 [tex]e ^{\lambda \ x}[/tex] = 0

[tex]e ^{\lambda \ x} (\lambda ^3 + 9 \lambda ^2 + \lambda - 111 )= 0[/tex]

∴

In as much as [tex]e^{ \lambda x}\neq 0[/tex] for any finite [tex]\lambda[/tex]; Then:

[tex]\lambda ^3 + 9 \lambda ^2 + 111 = 0[/tex]

By Factorization:

[tex](\lambda - 3) ( \lambda ^2 + 12 \lambda + 37) = 0[/tex]

[tex]\lambda = -6 + i \ or\ \lambda = -6 - i \ or \ \lambda = 3[/tex]

However;

The root [tex]\lambda = -6 \pm i[/tex] yield;

[tex]y_1 = (x) = c_1 e ^ {(-6+i)x}[/tex]

[tex]y_2 (x) = c_2e^{(-6-i)x}[/tex]

The root [tex]\lambda = 3[/tex] yield;

[tex]y_3(x) = c_3 e^{3x}[/tex]

∴

The general solution is:

[tex]y(x) = y_1(x) + y_2(x) + y_3(x) = \dfrac{c_1}{c^{(6-i)}x}+\dfrac{c_2}{c^{(6+i)}x}+ c_3e^{3x}[/tex]

Using Euler's Identity ;

[tex]e^{\alpha+i \beta} = e^\alpha \ cos (\beta ) + i \ e^\alpha \ sin ( \beta)[/tex]

[tex]y(x) = c_1 ( \dfrac{cos (x) }{e^{6x}}+ \dfrac{i \ sin x }{e^{6x} }) + c_2 ( \dfrac{cos (x)}{e^{6x}}- \dfrac{-i \ sin (x)}{e^{6x}})+c_3 e^{3x}[/tex]

[tex]\mathbf{y(x) = \dfrac{(c_1 + c_2) \ cos (x)}{e^{6x} }+ \dfrac{i(c_1 -c_2) \ sin (x)}{e^{6x}}+ c_3e^{3\ x}}[/tex]

Related Questions

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m = 5? Repeat these calculations for the cases where m = 10, m = 20, m = 30, and m = 50.

Required:
Discuss the results that you have obtained.

Answers

Answer:

The solution to the issue is outlined in the following portion of the summary.

Step-by-step explanation:

The given value is:

p = 0.0027

When m = 5,