Answer:
The largest interval is [tex]-\infty < 0 < 5[/tex]
Step-by-step explanation:
From the question the equation given is
[tex](x-5)y'' + 3y = x \ \ \ y(0) = 0 \ , y'(0) = 1[/tex]
Now dividing the both sides of this equation by (x-5)
[tex]y'' + \frac{3y}{(x-5)} = \frac{x}{x-5}[/tex]
Comparing this equation with the standard form of 2nd degree differential which is
[tex]y'' + P(x)y' + Q(x) y = R(x)[/tex]
We see that
[tex]Q(x) = \frac{3y}{(x-5)}[/tex]
[tex]R(x) = \frac{x}{(x-5)}[/tex]
So at x = 5 [tex]Q(x) \ and \ R(x)[/tex] are defined for this equation because from the equation of [tex]Q(x) \ and \ R(x)[/tex] x = 5 give infinity
This implies that the largest interval which includes x = 0 , P(x) , Q(x) , R(x ) is
[tex]-\infty < 0 < 5[/tex]
This because x = 5 is not defined in y domain
