# Determine the longest interval in which the given initial value

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.
$\left(x-2\right)y{}^{″}+{y}^{\prime }+\left(x-2\right)\left(\mathrm{tan}x\right)y=0$
$y\left(3\right)=1$
${y}^{\prime }\left(3\right)=6$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Marian Tucker
The given IVP
$\left(x-2\right)y{}^{″}+{y}^{\prime }+\left(x-2\right)\left(\mathrm{tan}x\right)y=0$
$⇒y{}^{″}+\frac{1}{x-2}{y}^{\prime }+\left(\mathrm{tan}x\right)y=0$
And the initial condition.
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: $\left(\left(2n+1\right)\frac{\pi }{2},\left(2n+3\right)\frac{\pi }{2}\right)$