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

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

Explanation:

The objective of this question is to solve:

`(dy(x))/(dx)+ 9(d^2y(dx))/(dx^2)+(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.

`(d^3)/(dx^3)(e ^(\lambda x) )+ 9 (d^2)/(dx^2)(e ^(\lambda x) ) + (d)/(dx)(e ^(\lambda x) )- 111 \ e ^(\lambda x) = 0`

To Replace:

`(d^3)/(dx^3)(e ^(\lambda x) )` with
`\lambda^3 e ^(\lambda x )`

`(d^2)/(dx^2)(e ^(\lambda x) ) \ with \ \lambda^2 e^(\lambda\ x)`

`(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)\\eq 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) = (c_1)/(c^((6-i))x)+(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 ( (cos (x) )/(e^(6x))+ (i \ sin x )/(e^(6x) )) + c_2 ( (cos (x))/(e^(6x))- (-i \ sin (x))/(e^(6x)))+c_3 e^(3x)`

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