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2nd order differential equation quick question http://tnypic.net/73979.jpg
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Just wondering if I just factorize this out?
\[x_1=e^{t/2}\\ x_1'=\frac{1}{2}e^{t/2}\\ x_1''=\frac{1}{4}e^{t/2}\\ 4x_1''-4x_1'+x=e^{t/2}-2e^{t/2}+e^{t/2}=0\] Thus, x_1 is a solution.
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