# 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

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