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

Wribreeminsl

Wribreeminsl

Answered question

2021-01-10

Using the second order linear ordinary differential equations existence and uniqueness theorem , find the largest interval in which the solution to the initial value is certain to exist.
t(t24)yty+3t2y=0,y(1)=1y(1)=3

Answer & Explanation

2abehn

2abehn

Skilled2021-01-11Added 88 answers

The basic value issue presented is:
t(t24)yty+3t2y=0,y(1)=1y(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(x0)=y0,y(x0)=y0
has a unique solution defined for all x in [a,b].
The differential equation with the following definition. ytt(t24)y+3t2t(t24)y=0
ytt24y+3t2t24y=0
Compare the above equation to the typical starting value problem.
Then, p(t)=1t24,q(t)=3tt24 and r(t)=0,t0=1 and y0=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 (,2)(2,2)(2,)
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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