The differential equation of the family of the curves y=ex(Acosx+Bsinx), where A and B are arbitrary constants is
A
d2ydx2+(dydx)2+y=0
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B
d2ydx2−2dydx+2y=0
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C
d2ydx2+2dydx−2y=0
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D
d2ydx2−2dydx+y=0
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Solution
The correct option is Bd2ydx2−2dydx+2y=0 ∵y=ex(Acosx+Bsinx) -------(i) ⇒dydx=ex(Acosx+Bsinx)+ex(−Asinx+Bcosx) ⇒dydx=y+ex(−Asinx+Bcosx) -------(ii)
Again differentiating both sides w.r.t. x, we get d2ydx2=dydx+ex(−Asinx+Bcosx)+ex(−Acosx−Bsinx) ⇒d2ydx2=dydx+dydx−y−y ∴d2ydx2−2dydx+2y=0