MHB Converting a Second-Order IVP into a System of Equations: Can Substitution Help?

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Change the second-order IVP into a system of equations
$y''+y'-2y=0 \quad y(0)= 2\quad y'(0)=0$

let $u=y'$

ok I stuck on this substitution stuff
 
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Since y'= u, y''= u' so y''+ y'- 2y= u'+ u- 2y= 0 so u'= 2y-u.
 

Thread 'A question about variables of integration'

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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