# Existence of a unique solution with given initial value problems.

Directions: Find an interval centered about $x = 0$ for which the given initial-value problem has a unique solution.

$$(x - 2)y'' + 3y = x$$

Initial values: $y(0) = 0,\,\,y'(0) = 1$.

My answer was $(-\infty, 1) \cup (3, \infty)$.

The books answer was $(-\infty < x < 2)$.

I determined my answer because I thought as long as

$a_{2}(x)$(the coefficient of the highest order derivative) $= 0$ is not true.

Maybe it's because I'm not really sure what "centered around $x = 0$" means. But can someone explain to me why the books answer is correct and mine is incorrect.

• Divide your initial value problem by $(x-2)$ and you will see why. Another hint, the initial conditions are $y(0) = 1$ and $y^{\prime}(0) = 1$ what does that tell your about your interval?
– Lays
Apr 23, 2013 at 4:07
• Also, you should accept your other questions answers :). "You can accept an answer by clicking on the ✓ to the left of the answer you'd like to accept."
– Lays
Apr 23, 2013 at 4:14
• I'm not sure what dividing by (x - 2) does. It just turns into $y^{''}$ + $\frac{3y}{x-2} = \frac{x}{x-2}$. I'm not sure how I can use the initial conditions to help determine the interval. Apr 23, 2013 at 18:11
• Where are the discontinuous points?
– Lays
Apr 23, 2013 at 19:55
• @Lays When x = 2. Apr 23, 2013 at 21:14

Ok, so the book states $a_2(x)$ not equal to $0$. So if you set $x = 2$ , it is the only condition where you will have a non unique solution. That gives the interval $(-\infty,2) \cup (2,\infty)$. However we need to choose an interval containing $x =0$. The only interval that does that is $x$ belongs to $(-\infty, 2)$. Hope it helped!