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4vir4ik [10]
3 years ago
13

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 − 5)y'' + 3y = x, y(0) = 0, y'(0) = 1
Mathematics
1 answer:
gavmur [86]3 years ago
4 0

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

       

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