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

The given initial value problem is:
$t(t^2-4)y^{″}-t{y}^{\prime }+3t^{2}y=0,y(1)=1y^{\prime }(1)=3$
If p(x), q(x) and g(x) are continuous on the interval [a,b], then the second order differential equation
$y^{″}+p(x)y^{\prime }+q(x)y=g(x),y(x_0)=y_0,y^{\prime }(x_0)=y_0^{\prime }$
has a unique solution defined for all x in [a,b].
Redefine the differential equation as follows.$y^{″}-\frac{t}{t(t^2-4)}{y}^{\prime }+\frac{3t^2}{t(t^2-4)}y=0$
$y^{″}-\frac{t}{t^2-4}{y}^{\prime }+\frac{3t^2}{t^2-4}y=0$
Compare the above equation with standard equation of initial value problem.
Then, $p(t{)}^{\prime }=\frac{1}{t^2-4},q(t)=\frac{3t}{t^2-4}$ 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 $(-\mathrm{\infty },-2)\cup (-2,2)\cup (2,\mathrm{\infty })$
Thus, the function has a unique solution at the point t=1 on the largest interval (-2,2)
Answer:
The solution to the initial value problem exist on the largest interval is (−2, 2).