dydx+1+y2x=1+y22etan−1y I.F. =etan−1yxetan−1y=∫1+y22(etan−1y)2dy Put tan−1y=t,1+y2dy=dtxetan−1y=∫2e2tdtxetan−1y=e2tan−1y+cx=etan−1y+ce−tan−1y∵y=0,x=11=1+c⇒c=0y=31,x=eπ/6
If x=f(y) is the solution of the differential equation (1+y2)+(x−2etan−1y) dxdy=0,y∈(−2π,2π) with f(0)=1, then f(31) is equal to :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
eπ/12
eπ/4
eπ/3
eπ/6
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