The given initial value problem is:

$t({t}^{2}-4){y}^{\u2033}-t{y}^{\prime}+3{t}^{2}y=0,y(1)=1{y}^{\prime}(1)=3$

If p(x), q(x) and g(x) are continuous on the interval [a,b], then the second order differential equation

${y}^{\u2033}+p(x){y}^{\prime}+q(x)y=g(x),y({x}_{0})={y}_{0},{y}^{\prime}({x}_{0})={y}_{0}^{\prime}$

has a unique solution defined for all x in [a,b].

Redefine the differential equation as follows.${y}^{\u2033}-\frac{t}{t({t}^{2}-4)}{y}^{\prime}+\frac{3{t}^{2}}{t({t}^{2}-4)}y=0$

${y}^{\u2033}-\frac{t}{{t}^{2}-4}{y}^{\prime}+\frac{3{t}^{2}}{{t}^{2}-4}y=0$

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).

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