- #1
shamieh
- 539
- 0
Find the explicit solution to $tyy' − 1 = 0$, $y(1) = 4$ and give the interval
of validity.
$ty \frac{dy}{dt} - 1 = 0$
$y \frac{dy}{dt} - 1 = 0$ ==> $ydy - {1/t} dt = 0$
$ydy = 1/t dt$
$\frac{y^2}{2} = ln(t) + c$$y = \sqrt{2ln(t) + c}$
applying $y(1) = 4$
so the explicit solution is:
$C = 4$
But I'm not sure how to get the interval of validity?
of validity.
$ty \frac{dy}{dt} - 1 = 0$
$y \frac{dy}{dt} - 1 = 0$ ==> $ydy - {1/t} dt = 0$
$ydy = 1/t dt$
$\frac{y^2}{2} = ln(t) + c$$y = \sqrt{2ln(t) + c}$
applying $y(1) = 4$
so the explicit solution is:
$C = 4$
But I'm not sure how to get the interval of validity?