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: <math>\lambda^2+4\lambda+4=(\lambda+2)^2=0 </math>
Since <math>\lambda=-2</math> is a repeated root, we have to introduce a factor of ''x'' for one solution to ensure linear independence: <math> u_1 = e^{-2x} </math> and <math> u_2 =x e^{-2x}</math>. The [[Wronskian]] of these two functions is
: <math>W=\begin{vmatrix}
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