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Question Number 183600 by Mastermind last updated on 27/Dec/22

y′^(′′)  + 8y′^(′′)  +12y′ = 0    Solve with better explanation

$$\mathrm{y}'^{''} \:+\:\mathrm{8y}'^{''} \:+\mathrm{12y}'\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Solve}\:\mathrm{with}\:\mathrm{better}\:\mathrm{explanation} \\ $$

Commented by mr W last updated on 27/Dec/22

you have caused other people doing  unnecessary work to solve your  questions which you didn′t mean,  because you had typos, like Q183209.  so please be more careful when posting  a question, check if you have typos  before sending!  i think you have here typos again:  y′^(′′)  + 8y′^(′′)  +12y′ = 0    let other people do unnecessary  work is also disrespectful!  making mistake is excusable, but  making the same mistake repeatedly  is inexcusable!

$${you}\:{have}\:{caused}\:{other}\:{people}\:{doing} \\ $$$${unnecessary}\:{work}\:{to}\:{solve}\:{your} \\ $$$${questions}\:{which}\:{you}\:{didn}'{t}\:{mean}, \\ $$$${because}\:{you}\:{had}\:{typos},\:{like}\:{Q}\mathrm{183209}. \\ $$$${so}\:{please}\:{be}\:{more}\:{careful}\:{when}\:{posting} \\ $$$${a}\:{question},\:{check}\:{if}\:{you}\:{have}\:{typos} \\ $$$${before}\:{sending}! \\ $$$${i}\:{think}\:{you}\:{have}\:{here}\:{typos}\:{again}: \\ $$$$\mathrm{y}'^{''} \:+\:\mathrm{8y}'^{''} \:+\mathrm{12y}'\:=\:\mathrm{0} \\ $$$$ \\ $$$${let}\:{other}\:{people}\:{do}\:{unnecessary} \\ $$$${work}\:{is}\:{also}\:{disrespectful}! \\ $$$${making}\:{mistake}\:{is}\:{excusable},\:{but} \\ $$$${making}\:{the}\:{same}\:{mistake}\:{repeatedly} \\ $$$${is}\:{inexcusable}! \\ $$

Commented by Mastermind last updated on 27/Dec/22

I understand you BOSS, thank you

$$\mathrm{I}\:\mathrm{understand}\:\mathrm{you}\:\mathrm{BOSS},\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mr W last updated on 27/Dec/22

then please make your question clear!  9y′^(′′)  +12y′ = 0  or  y′^(′′)  + 8y′′ +12y′ = 0

$${then}\:{please}\:{make}\:{your}\:{question}\:{clear}! \\ $$$$\mathrm{9y}'^{''} \:+\mathrm{12y}'\:=\:\mathrm{0} \\ $$$${or} \\ $$$$\mathrm{y}'^{''} \:+\:\mathrm{8y}''\:+\mathrm{12y}'\:=\:\mathrm{0} \\ $$

Answered by CElcedricjunior last updated on 27/Dec/22

y(x)=ae^(−6x) +be^(−2x) +c    a;bet c∈R

$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ae}}^{−\mathrm{6}\boldsymbol{{x}}} +\boldsymbol{{be}}^{−\mathrm{2}\boldsymbol{{x}}} +\boldsymbol{{c}}\:\: \\ $$$$\boldsymbol{{a}};\boldsymbol{{bet}}\:\boldsymbol{{c}}\in\mathbb{R} \\ $$$$ \\ $$

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