verskalksv
2021-11-15
Answered

Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the slotuion.

$(x-2)y{}^{\u2033}+{y}^{\prime}+(x-2)\left(\mathrm{tan}x\right)y=0$

$y\left(3\right)=1$

${y}^{\prime}\left(3\right)=6$

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Marian Tucker

Answered 2021-11-16
Author has **15** answers

The given IVP

$(x-2)y{}^{\u2033}+{y}^{\prime}+(x-2)\left(\mathrm{tan}x\right)y=0$

$\Rightarrow y{}^{\u2033}+\frac{1}{x-2}{y}^{\prime}+\left(\mathrm{tan}x\right)y=0$

And the initial condition.$y\left(3\right)=1,\text{}{y}^{\prime}\left(3\right)=6$

Here$P\left(x\right)=\frac{1}{x-2}$

$q\left(x\right)=\mathrm{tan}x$

$g\left(x\right)=0$ constant function is continuous everywhere.

Solution:$((2n+1)\frac{\pi}{2},(2n+3)\frac{\pi}{2})$

And the initial condition.

Here

Solution:

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