Question

My Notes 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 solution. (Enter your answer using interval notation.)t(t−4)y"+3ty'+4y=2,y(3)=0,y'(3)=−1

Answer 1

Answer:

The answer to the question is

The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is  (-∞, 4)

Step-by-step explanation:

To apply look for the interval, we divide the ordinary differential equation by (t-4) to

y'' + $\frac{3t}{t-4}$ y' + $\frac{4}{t-4}$y = $\frac{2}{t-4}$

Using theorem 3.2.1 we have p(t) =  $\frac{3t}{t-4}$, q(t) =  $\frac{4}{t-4}$, g(t) = $\frac{2}{t-4}$

Which are undefined at 4. Therefore the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution, that is where p, q and g are continuous and defined is (-∞, 4) whereby theorem 3.2.1 guarantees unique solution satisfying the initial value problem in this interval.