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(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.
\(t(t^2-4)y''-ty'+3t^2y=0, y(1)=1 y'(1)=3\)

Answers (1)

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)
Answer:
The solution to the initial value problem exist on the largest interval is (−2, 2).
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