# Using the existence and uniqueness theorem for second order linear ordinary differential equations, find the largest interval in which the solution to the initial value is certain to exist. t(t^2-4)y''-ty'+3t^2y=0, y(1)=1 y'(1)=3

Question
Using the existence and uniqueness theorem for second order linear ordinary differential equations, find the largest interval in which the solution to the initial value is certain to exist.
$$t(t^2-4)y''-ty'+3t^2y=0, y(1)=1 y'(1)=3$$

2021-01-11
The given initial value problem is:
$$t(t^2-4)y''-ty'+3t^2y=0, y(1)=1 y'(1)=3$$
If p(x), q(x) and g(x) are continuous on the interval [a,b], then the second order differential equation
$$y''+p(x)y'+q(x)y=g(x), y(x_0)=y_0, y'(x_0)=y_0'$$
has a unique solution defined for all x in [a,b].
Redefine the differential equation as follows. $$y''-\frac{t}{t(t^2-4)}y'+\frac{3t^2}{t(t^2-4)}y=0$$
$$y''-\frac{t}{t^2-4}y'+\frac{3t^2}{t^2-4}y=0$$
Compare the above equation with standard equation of initial value problem.
Then, $$p(t)'=\frac{1}{t^2-4},q(t)=\frac{3t}{t^2-4}$$ and $$r(t)=0, t_0=1$$ and $$y_0=1$$
The function p(t) is continuous for all values of t except t=w and t=-2
The function q(t) is continuous for all values of t except t=w and t=-2
The domain of the both the function is $$(-\infty,-2)\cup(-2,2)\cup(2,\infty)$$
Thus, the function has a unique solution at the point t=1 on the largest interval (-2,2)
The solution to the initial value problem exist on the largest interval is (−2, 2).

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