MHB Finding interval where second order ODE has unique solution

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The discussion revolves around applying the existence and uniqueness theorem to the second-order ordinary differential equation (ODE) given by y'' + tan(x)y = e^x, with initial conditions y(0) = 1 and y'(0) = 0. The user reformulates the ODE into a standard form, identifying a_2(x) = cos(x), which is non-zero at x = 0, thus satisfying the conditions of the theorem. This confirms that a unique solution exists in an interval around x = 0. The focus is on understanding how to properly apply the theorem to establish the existence of solutions for the given initial value problem. The discussion emphasizes the importance of verifying the conditions of the theorem to ensure a unique solution exists.
find_the_fun
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I'm a little stuck getting started on this question. $$y''+\tan(x)y=e^x$$ with $$y(0)=1,y'(0)=0$$. I know the existence and uniqueness theorem
Let $$a_n(x),a_{n-1}(x),...,a_0(x)$$ and g(x) be continuous on an Interval I and let $$a_n(x)$$ not be 0 for every x in this interval. If x=x_0 is any point in this interval, then a solution Y(x) of the initial value problem exists on the interval and is unique.
for an nth order initial value problem

How do I apply the theorem?
 
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find_the_fun said:
I'm a little stuck getting started on this question. $$y''+\tan(x)y=e^x$$ with $$y(0)=1,y'(0)=0$$. I know the existence and uniqueness theorem
for an nth order initial value problem

How do I apply the theorem?

Let's write the second order linear ODE as...

$\displaystyle \cos x\ y^{\ ''} + \sin x\ y = e^{x}\ \cos x, \ y(0)=1,\ y^{\ '} (0)=0\ (1)$

Here $a_{2}(x) = \cos x$, and is $a_{2} (0) = 1 \ne 0$, so that the existence and uniqueness theorem is verified...

Kind regards

$\chi$ $\sigma$
 

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