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Question

# The equation of the curve passing through the origin and satisfying the differential equation (dydx)2=(x−y)2, is

A
e2x(1x+y)=1+xy
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B
e2x(1+xy)=1x+y
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C
e2x(1x+y)=1+x+y
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D
e2x(1+x+y)=1x+y
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Solution

## The correct option is A e2x(1−x+y)=1+x−yWe have, dydx=±(x−y) Case 1 When dydx=(x−y) In this case, we have 1−dvdx=v, where x−y=v⇒dvdx=1−v⇒11−vdv=dx⇒−log(1−v)=x+logc⇒(1−v)−1=cex⇒11−x+y=cex It passes through the origin. ∴ c=1Hence, 11−x+y=ex .....(i) Case 2 When dydx=−(x−y)=y−x In this case, we have dudx+1=u, where u=y−x ⇒duu−1=dx⇒log(u−1)=x+logc⇒u−1=cex⇒y−x−1=cex It passes through the origin. ∴ c=−1Hence, y−x−1=−ex⇒x−y+1=ex From Eqs. (i) and (ii), we obtain x−y+11−x+y=e2x⇒ (x−y+1)=(1−x+y)e2x

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