MHB Finding interval where second order ODE has unique solution

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

$\chi$ $\sigma$
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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