I have to determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.

Question

The equation is the following: $$(t-6)y' + (\ln t)y = 4t,~~~~~~ y(1) = 4$$

I really tried do understand how to do it but I failed and there is nothing about it in my notebook so any help is appreciated. Thank you.

Answer 1

The Solution will exist in $t\in (0,6).$ because $\ln 0= -\infty,$ and $t=6$ the ODE does not exist. Since $y(1)$ is given so $t \in (6,\infty)$ cannot be considered.

Answer 2

Just find the intervals where the coefficients and right side of the normalized equation $$ y'+\frac{\ln t}{t-6}y=\frac{4t}{t-6} $$ exist and are all continuous. Then pick the one that contains $t=1$.

Answer 3

Hint: I would first put the ODE in standard form

$$\frac{dy}{dt} + P(t)y = Q(t)$$

which in your case is

$$\frac{dy}{dt} + \frac{\ln t}{t-6}y = \frac{4t}{t-6}$$

and then observe the intervals in which $P(t)$ and $Q(t)$ are continuous. Note that both $P(t)$ and $Q(t)$ are undefined at $t=6$.

After you find the intervals of continuity for $P(t)$ and $Q(t)$, combine them together in one interval in which both $P(t)$ and $Q(t)$ are continuous.

Then confirm the initial condition at $t=1$ is in the interval of continuity. In your case, $y(1)=4$ should be in the interval of continuity for $P(t)$ and $Q(t)$.