Solve the given LDE: y6−64y=0

Name: Solve the given LDE: y6−64y=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the given LDE: y6−64y=0

Question

Archie Jones · Accepted Answer

In this question we will use basic of calculus i.e concept of differential equations.
We will simply form the auxiliary equation and then we will find the roots of that equation after that we will write the solution of differential equation.
Solution of the above question is in the step below.

y^{6} - 64 y = 0

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

where

D = \frac{d}{d x}

m^{6} - 64 = 0

{(m^{2})}^{3} - {(4)}^{3} = 0

{(m^{3})}^{2} - {(8)}^{2} = 0

(m^{3} + 8) (m^{3} - 8) = 0

(m + 2) (m^{2} - 2 m + 4) (m - 2) (m^{2} + 2 m + 4) = 0

So,

m = - 2, - 2, - 2, 2, 2, 2

.
So, m has two roots and supeated there times each.
Hence, solution of differential equation is

y = (c_{1} + c_{2} x + c_{3} x^{2}) e^{- 2 x} + (c_{4} + c_{5} x + c_{6} x^{2}) e^{2 x}

Hattie Schaeffer · Accepted Answer

Given that the differential equation,  y6−64y=0  Solve it.  Solution: Given, y6−64y=0 (1)  Let y=Aemx be the trial solution (1)  Then y6=Am6emx [y(6) means, 6th order derivative]  So from (1), Am6emx−64Aemx=0  ⇒Aemx(m6−64)=0  ⇒m6−64=0 (Since Aemx≠0)  This is an auxiliary equation.  Now Solve m6−64=0  ⇒(m3)2−(8)2=0  ⇒(m3+8)(m3−8)=0  ⇒(m3+23)(m3−23)=0  ⇒{(m+2)(m2−2m+4)}{(m−2)(m2+2m+4)}=0  ⇒(m+2)(m−2)(m2−2m+4)(m2+2m+4)=0  So m+2=0⇒m=−2  m−2=0⇒m=2  m2−2m+4=0⇒m=−(−2)±(−2)2−4⋅1⋅42⋅1  And m2+2m+4=0  ⇒m=−2±(2)2−9⋅1⋅42⋅1  =−2±4−162  =−2±−122  =−2±2−32=−1±−3=−1±i3

RizerMix · Accepted Answer

y6−64y=0 We add all the numbers together, and all the variables -64y=0 y=0/-64 Y=

Solve the given LDE: y^{6} - 64y=0

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