Question

Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answer using interval notation.) (x − 3)y'' + 4y = x, y(0) = 0, y'(0) = 1

Answer 1

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=$(-\infty,3)$