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dbhuggybearow6jng

18.02.2020 •

Mathematics

Consider the initial value problem y' + 2 3 y = 1 − t 2 , y(0) = y0. Find the value of y0 for which the solution touches, but does not cross, the t-axis. (A computer algebra system is recommended. Round your answer to three decimal places.)

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Ответ:

lily1482

18.01.2022

$y_\circ= 28.911$

Step-by-step explanation:

Given,

$y'+\frac{2}{3}y=1-\frac{1}{2} t$

Integrating factor= $e^{\int{\frac{2}{3} dt}}=e^{\frac{2}{3}t}$

Multiplying the integrating factor

$e^{\frac{2}{3}t} y'+\frac{2}{3}e^{\frac{2}{3}t}}y=e^{\frac{2}{3}t}}-\frac{1}{2} e^{\frac{2}{3}t}}t$

Integrating both sides

$\int (e^{\frac{2}{3}t} y'+\frac{2}{3}e^{\frac{2}{3}t}}y)dt=\int (e^{\frac{2}{3}t}}-\frac{1}{2} e^{\frac{2}{3}t}}t)dt$ Let $I=\int e^{\frac{2}{3}t}t dt$

$\Rightarrow e^{\frac{2}{3}t}}y=\frac{21}{8}e^{\frac{2}{3}t}}-\frac{3}{4} e^{\frac{2}{3}t}}t+C$ $=t\int e^{\frac{2}{3}t}dt - \int[\int e^{\frac{2}{3}t}dt\times \frac{d}{dt} (t)]dt$

$=\frac{3te^{\frac{2}{3}t} }{2} -\frac{9}{4} e^{\frac{2}{3}t} }$

Given,y(0)=y₀

$\Rightarrow e^{\frac{2}{3}t}}y_\circ=\frac{21}{8}e^{\frac{2}{3}\times 0}}-\frac{3}{4} e^{\frac{2}{3} \times 0}}\times 0+C$

$\Rightarrow C=e^{\frac{2}{3}t}}y_\circ-\frac{21}{8}$

$\therefore e^{\frac{2}{3}t}}y=\frac{21}{8}e^{\frac{2}{3}t}}-\frac{3}{4} e^{\frac{2}{3}t}}t+e^{\frac{2}{3}t}}y_\circ-\frac{21}{8}$

y(t)=0

$\therefore 0 =\frac{21}{8}e^{\frac{2}{3}t}}-\frac{3}{4} e^{\frac{2}{3}t}}t+e^{\frac{2}{3}t}}y_\circ-\frac{21}{8}$

$(y_\circ-\frac{21}{8} )e^{\frac{-2}{3} t}=\frac{3}{4}t -\frac{21}{8}$

Again since this solution has slope zero when it is just touche the t-axis.

y'(t)=0

$(y_\circ-\frac{21}{8} )e^{\frac{-2}{3} t}=\frac{9}{8}$

$\Rightarrow \frac{3}{4}t -\frac{21}{8}=\frac{9}{8}$ $[(y_\circ-\frac{21}{8} )e^{\frac{-2}{3} t}=\frac{3}{4}t -\frac{21}{8}]$

$\Rightarrow t=5$

Therefore

$(y_\circ-\frac{21}{8} )e^{\frac{-2}{3} 5}=\frac{3}{4}\times5 -\frac{21}{8}$

$\Rightarrow y_\circ= 28.911$

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Ответ:

apple1367

17.11.2022

It will be b a c that order least to greatest

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