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Question Number 103969 by bobhans last updated on 18/Jul/20
how do you solve (D^3 +12D^2 +36D)y=0  by constant coefficients
$${how}\:{do}\:{you}\:{solve}\:\left({D}^{\mathrm{3}} +\mathrm{12}{D}^{\mathrm{2}} +\mathrm{36}{D}\right){y}=\mathrm{0} \\ $$$${by}\:{constant}\:{coefficients} \\ $$
Answered by mr W last updated on 18/Jul/20
λ(λ+6)^2 =0  λ=0,−6,−6  ⇒y=(A+Bx)e^(−6x) +C
$$\lambda\left(\lambda+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\lambda=\mathrm{0},−\mathrm{6},−\mathrm{6} \\ $$$$\Rightarrow{y}=\left({A}+{Bx}\right){e}^{−\mathrm{6}{x}} +{C} \\ $$
Commented by bobhans last updated on 18/Jul/20
simple way
$${simple}\:{way} \\ $$

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