Question

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

Answer 1

Answer:

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

Step-by-step explanation:

The objective of this question is to solve:

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

Suppose the general solution is proportional to $e^{\lambda x}$ for $\lambda$ is constant; Then:

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

i.e.

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

To Replace:

$\dfrac{d^3}{dx^3}(e ^{\lambda x} )$ with $\lambda^3 e ^{\lambda x }$

$\dfrac{d^2}{dx^2}(e ^{\lambda x} ) \ with \ \lambda^2 e^{\lambda\ x}$

$\dfrac{d}{dx}(e ^{\lambda x} ) \ with \ \lambda e ^{\lambda \ x}$

Thus;

$\lambda^3 e ^{\lambda x }$ + $9 \lambda^2 e^{\lambda\ x}$ + $\lambda e ^{\lambda \ x}$ - 111 $e ^{\lambda \ x}$ = 0

$e ^{\lambda \ x} (\lambda ^3 + 9 \lambda ^2 + \lambda - 111 )= 0$

∴

In as much as $e^{ \lambda x}\neq 0$ for any finite $\lambda$; Then:

$\lambda ^3 + 9 \lambda ^2 + 111 = 0$

By Factorization:

$(\lambda - 3) ( \lambda ^2 + 12 \lambda + 37) = 0$

$\lambda = -6 + i \ or\ \lambda = -6 - i \ or \ \lambda = 3$

However;

The root $\lambda = -6 \pm i$ yield;

$y_1 = (x) = c_1 e ^ {(-6+i)x}$

$y_2 (x) = c_2e^{(-6-i)x}$

The root $\lambda = 3$ yield;

$y_3(x) = c_3 e^{3x}$

∴

The general solution is:

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

Using Euler's Identity ;

$e^{\alpha+i \beta} = e^\alpha \ cos (\beta ) + i \ e^\alpha \ sin ( \beta)$

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

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