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Name: Solve the following linear differential equation y6−64y=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the following linear differential equation y6−64y=0

Stacie Worsley

Answered question

2021-12-17

Solve the following linear differential equation

y^{6} - 64 y = 0

Answer & Explanation

esfloravaou

Beginner2021-12-18Added 43 answers

y^{6} - 64 y = 0

\frac{d^{6} y}{{d x}^{6}} - 64 x = 0

(D^{6} - 64) y = 0

So for C.F.

D^{6} - 64 = 0

D^{6} = 64

D^{6} = 2^{6} D

D = \pm 2

y = c_{1} e 6 {2 x} + c_{2} e^{- 2 x}

Barbara Meeker

Beginner2021-12-19Added 38 answers

So,

y^{6} - 64 y = 0

y = e^{m x}

is a solution

y^{1} = m e^{m x}

y^{6} = m^{6} e^{m x}

So,

m^{6} e^{m x} - 64 e^{m x} = 0

m^{6} = 64

So,

m = \pm 2, \pm (1 + i \sqrt{3}), \pm (- 1 + i \sqrt{3})

So,

m = - 2, + 2, i + i \sqrt{3}, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}, 1 - i \sqrt{3}

So,

[y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} e^{(1 + i \sqrt{3})} + c_{4} e^{- (1 + i \sqrt{3}) t} + c_{5} e^{(- 1 + i \sqrt{3}) t} + c_{6} e^{- (- 1 + i \sqrt{3}) i}]

RizerMix

Expert2021-12-29Added 656 answers

Pull out like factors:
$y^{6} - 64 y = y \cdot (y^{5} - 64)$
$y \cdot (y^{5} - 64) = 0$
A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term $= 0$ separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term $= 0$ solves product $= 0$ as well.
Solve: $y = 0$
Solution is $y = 0$
Solve: $y^{5} - 64 = 0$
Add 64 to both sides of the equation :
$y^{5} = 64$
$y = 5 t h$ root of (64)
The equation has one real solution
This solution is $y = 2 \cdot 5 t h$ root of $2 = 2.2974$

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