Q.

he solution curve of the differential equation, 1+e-x1+y2dydx=y2 , which passes through the point (0,1), is :

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a

y2+1=yloge1+e−x2+2

b

y2=1+yloge1+ex2

c

y2+1=yloge1+ex2+2

d

y2=1+yloge1+e−x2

answer is B.

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Detailed Solution

Given equation is 1+y2y2dy=ex1+exdx    variables separable method⇒∫1y2+1dy=∫exdx1+ex⇒-1y+y=ln1+ex+C It passes through (0,1)-1+1=ln2+C⇒C=-ln2 Equation of curve y-1y=ln1+ex-ln2⇒y2-1=yln1+ex2
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