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Question: If \(y = x^{2}e^{x},\) then value of \(y_{n}\) is...

If y=x2ex,y = x^{2}e^{x}, then value of yny_{n} is

A

{x2−2nx+n(n−1)}ex\{ x^{2} - 2nx + n(n - 1)\} e^{x}

B

{x2+2nx+n(n−1)}ex\{ x^{2} + 2nx + n(n - 1)\} e^{x}

C

{x2+2nx−n(n−1)}ex\{ x^{2} + 2nx - n(n - 1)\} e^{x}

D

None of these

Answer

{x2+2nx+n(n−1)}ex\{ x^{2} + 2nx + n(n - 1)\} e^{x}

Explanation

Solution

Applying Leibnitz’s theorem by taking x2x^{2} as second function. We get , Dny=Dn(ex.x2)D^{n}y = D^{n}(e^{x}.x^{2})

= nC0Dn(ex)x2+n⥂C1Dn−1(ex).D(x2)+n⥂C2Dn−2(ex).D2(x2)+...........nC_{0}D^{n}(e^{x})x^{2} +^{n} ⥂ C_{1}D^{n - 1}(e^{x}).D(x^{2}) +^{n} ⥂ C_{2}D^{n - 2}(e^{x}).D^{2}(x^{2}) + ...........= ex.x2+nex.2x+n(n−1)2!ex.2+0+0+..........e^{x}.x^{2} + ne^{x}.2x + \frac{n(n - 1)}{2!}e^{x}.2 + 0 + 0 + ..........

yn={x2+2nx+n(n−1)}exy_{n} = \{ x^{2} + 2nx + n(n - 1)\} e^{x}.