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 − 5)y'' + 3y = x, y(0) = 0, y'(0) = 1

Answer 1

Answer:

The largest interval is    $-\infty < 0 < 5$

Step-by-step explanation:

From the question the equation given is  

       $(x-5)y'' + 3y = x \ \ \ y(0) = 0 \ , y'(0) = 1$  

Now dividing the both sides of this equation by (x-5)

         $y'' + \frac{3y}{(x-5)} = \frac{x}{x-5}$

Comparing this equation with the standard form of 2nd degree differential which is

        $y'' + P(x)y' + Q(x) y = R(x)$

We see that

        $Q(x) = \frac{3y}{(x-5)}$

        $R(x) = \frac{x}{(x-5)}$

So at x =  5  $Q(x) \ and \ R(x)$ are defined for this equation because from the equation of $Q(x) \ and \ R(x)$  x =  5 give infinity

This implies that the largest interval which includes x = 0 , P(x) , Q(x) , R(x ) is  

       $-\infty < 0 < 5$

This because x = 5 is not defined in y domain