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Solve y” – 2y’ + 2y = e^x tanx by variation of parameter.

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Solve y” – 2y’ + 2y = ex tanx by variation of parameter.

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Rewrite the given equation, (D2 – 2D + 2)y = extanx. 

Here A.E becomes, D2 – 2D + 2 = 0 D = 1 ± i

Thus, yC.F.(x) = c1y1 + c2y2 = ex[c1cosx + c2sinx] … (1) 

Assume: yP.I.(x) = u1(x)y1 + u2(x)y2 … (2) 

By method of variation of parameter

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