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y = Ae^(x) + Be^(-2x) is a solution of t...

` y = Ae^(x) + Be^(-2x)` is a solution of the D.E. ` (d^(2)y)/(dx^(2) )+ (dy)/(dx) -2y = 0`

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Differentiating ` y = Ae^(x) + Be^(-2x)`
twice w.r.t x , we get
` (dy)/(dx) =Ae^(x) - 2Be^(-2x)`
` and(d^(2)y)/(dx^(2)) =Ae^(x) + 4Be^(-2x)`
LHS = `(d^(2)y)/(dx^(2)) + (dy)/(dx) - 2y`
` = (Ae^(x) +4Be^(-2x)) + (Ae^(x) -2Be^(-2x) -2 (Ae^(x) +Be^(-2x))`
` = Ae^(x) + 4Be^(-2x) - 2Be^(-2x) -2Ae^(x) -2Be^(-2x)`
=0= RHS
The shows that ` y = Ae^(x) + Be^(-2x) ` is a solution of the D.E . ` (d^(2)y)/(dx^(2)) + (dy)/(dx) - 2y = 0 `
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