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Question

Find the differential equation of family of curves y=ex(Acosx+Bsinx) where A and B are arbitrary constants.

Solution
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y=ex(Acosx+Bsinx)
y1=ex(Asinx+Bcosx)+ex(Acosx+Bsinx)
y1=ex(A(cosxsinx)+B(cosx+sinx))
y2=ex(A(sinxcosx)+B(sinx+cosx))+ex(A(cosxsinx)+B(cosx+sinx))
y2=ex(A(2sinx))+B(2cosx))
y2=2ex(BcosxAsinx)
As, y+y22=y1
y22y1+2y=0

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