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ra1l [238]
3 years ago
10

Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answ

er using interval notation.) (x − 3)y'' + 4y = x, y(0) = 0, y'(0) = 1
Mathematics
1 answer:
Ivan3 years ago
7 0

Answer:

(-\infty,3)

Step-by-step explanation:

We are given that

(x-3)y''+4y=x

y''+\frac{4}{x-3}y=\frac{x}{x-3}

y(0)=0

y'(0)=1

By comparing with

y''+p(x)y'+q(x)y=g(x)

We get

p(x)=\frac{4}{x-3}

g(x)=\frac{x}{x-3}

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

(-\infty,3)and (3,\infty)

By unique existence theorem

Largest interval which contains 0=(-\infty,3)

Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=(-\infty,3)

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