Answer:
[tex](-\infty,3)[/tex]
Step-by-step explanation:
We are given that
[tex](x-3)y''+4y=x[/tex]
[tex]y''+\frac{4}{x-3}y=\frac{x}{x-3}[/tex]
y(0)=0
y'(0)=1
By comparing with
[tex]y''+p(x)y'+q(x)y=g(x)[/tex]
We get
[tex]p(x)=\frac{4}{x-3}[/tex]
[tex]g(x)=\frac{x}{x-3}[/tex]
q(x)=0
p(x),q(x) and g(x) are continuous for all real values of x except 3.
Interval on which p(x),q(x) and g(x) are continuous
[tex](-\infty,3)[/tex]and (3,[tex]\infty)[/tex]
By unique existence theorem
Largest interval which contains 0=[tex](-\infty,3)[/tex]
Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=[tex](-\infty,3)[/tex]
The largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]
The given parameters are:
(x − 3)y'' + 4y = x,
y(0) = 0
y'(0) = 1
Divide the equation (x − 3)y'' + 4y = x through by (x - 3)
[tex]y'' + \frac{4y}{x - 3} = \frac{x}{x - 3}[/tex]
Compare the above equation to the following equation
y" + p(x) y' + q(x)y = g(x)
Then, we have:
[tex]p(x) = \frac{4y}{x - 3}[/tex]
[tex]q(x) = 0[/tex]
[tex]g(x) = \frac x{x - 3}[/tex]
The domains of functions p(x) and g(x) are all set of real values except 3
This is represented as:
[tex](-\infty, 3)\ u\ (3,\infty)[/tex]
Using the unique existence theorem, we have:
The largest interval that contains x = 0 is [tex](-\infty, 3)[/tex]
Hence, the largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]
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