Solve Initial-Value Problem: Find Interval x=0

[KillerZ]

Homework Statement

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

$$y^{''} + (tanx)y = e^{x}$$

$$y(0) = 1$$ $$y^{'}(0) = 0$$

Homework Equations

$$a_{i}(x), i=0,1,2,3,...,n$$ is continuous and

$$a_{n} \neq 0$$ for every x in I.

The Attempt at a Solution

$$a_{0} = tanx$$ is zero at x = 0

I am not sure if this is correct because tanx is continuous everywhere except at pi/2, 3pi/2, etc... so would interval be:

$$I = (0,\infty) or (-\infty , 0)$$

or

$$I = (0,\pi/2) or (-\pi/2 , 0)$$

 

[Mark44]

Homework Statement

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

$$y^{''} + (tanx)y = e^{x}$$

$$y(0) = 1$$ $$y^{'}(0) = 0$$

Homework Equations

How are what you have below relevant? What does a_i(x) represent?

$$a_{i}(x), i=0,1,2,3,...,n$$ is continuous and

$$a_{n} \neq 0$$ for every x in I.

The Attempt at a Solution

$$a_{0} = tanx$$ is zero at x = 0

I am not sure if this is correct because tanx is continuous everywhere except at pi/2, 3pi/2, etc... so would interval be:

$$I = (0,\infty) or (-\infty , 0)$$

or

$$I = (0,\pi/2) or (-\pi/2 , 0)$$

Do you have a theorem that can be used for this problem? It might be titled Existence and Uniqueness Theorem.

 

[KillerZ]

I found the interval:

as tanx = sinx/cosx

cosx can not equal zero

so the interval is:

$$(-\frac{\pi}{2}, \frac{\pi}{2})$$