Question

Consider the DE y'' + 16y' + 64y = 0, which is linear with constant coefficients. First, we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is

          Consider the DE y'' + 16y' + 64y = 0, which is linear with constant coefficients. First, we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is
        

Added by Joseph G.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Consider the DE y'' + 16y' + 64y = 0, which is linear with constant coefficients. First, we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is
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Transcript

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00:01 Hi here for the given question.
00:03 We are given that here.
00:04 We have auxiliary equation as m multiplied with m plus 3 whole square multiplied with m square plus m plus 2 whole square is equal to 0.
00:15 So here for them we can write this as here.
00:19 We have value of m plus 3 equal to 0 m is equal to 0 which implies here.
00:26 We have e to the power 0 t which can be written as 1 for general solution.
00:31 So here we have m plus 3 is equal to 0.
00:34 So m is equal to minus 3 and it is whole square.
00:38 So repeated twice.
00:41 So here it can be written as e to the power minus 3.
00:45 Now further here we have m square plus m plus 2 is equal to 0.
00:52 So here m is equal to minus b plus or minus under root of b square minus 4 ac divided by 2a.
01:01 So substituting the values here we have minus 1 plus or minus under root of minus 7 divided by 2.
01:11 So here on simplifying further we have m is equal to minus 1 plus or minus under root 7 iota divided by 2...
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