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liubitsaghn2145

liubitsaghn2145

29.02.2020 •

Mathematics

Consider the following higher-order differential equation. d3u dt3 + d2u dt2 − 2u = 0 Find all the roots of the auxiliary equation. (Enter your answers as a comma-separated list.)

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Ответ:

priceisright11401

19.01.2022

Therefore the auxiliary solution is

$y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]$

Step-by-step explanation:

To find auxiliary equation we have to put $u=e^{mt}$ in the given differential equation.

The degree of the differential equation is 3.

Therefore the number of root of the differential equation is 3.

Let $\lambda_1$ , $\lambda_2$ and $\lambda_3$ be three roots of the auxiliary equation.

If three roots are real and equal.

Then $y= e^{\lambda_1t} (C_1+C_2t+C_3t^2)$

If three roots are real and distinct.

Then $y=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}+C_3e^{\lambda_3x}$

If two roots imaginary and one root real , $\lambda_1= a+ib \ and \ \lambda_2= a-ib$

Then $y=C_1e^{\lambda_3t}+e^{at}(C_2sin \ bt+C_3cos \ bt)$

Now, $u=e^{mt}$ , $u'=me^{mt},u"=m^2e^{mt} \ and \ u'"= m^3e^{mt}$

Given differential equation is

$\frac{d^3u}{dt^3} +\frac{d^2u}{dt^2}-2u=0$

The auxiliary equation is

$(m^3+m^2-2)e^{mt}=0$

$\Rightarrow (m^3+m^2-2)=0$

$\Rightarrow m^3-m^2+2m^2-2m+2m-2=0$

$\Rightarrow m^2(m-1)+2m(m-1)+2(m-1)=0$

$\Rightarrow (m-1)(m^2+2m+2)=0$

$\Rightarrow m= 1,\frac{-2\pm\sqrt{2^2-4.2.1} }{2.1}$

$\Rightarrow m= 1,\frac{-2\pm\sqrt{-4} }{2.1}$

$\Rightarrow m= 1,-1\pm i$

Here $\lambda_1= -1+i , \lambda_2=- 1-i \ and \ \lambda_3=1$

Therefore ,

$y=C_1e^{1.t}+e^{-1.t}[C_2sin \ (1.t)+C_3 cos \ (1.t)]$

$\Rightarrow y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]$

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Ответ:

joytheqt

07.04.2021

Step-by-step explanation:

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thanks for free point

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