Rewrite as dxdy+1+x2y=1+x2tan−1x.
IF =etan−1x.
y⋅etan−1x=∫1+x2tan−1x⋅etan−1xdx.
Let t=tan−1x: =∫tetdt=et(t−1)+C.
y=tan−1x−1+Ce−tan−1x.
Using y(0)=1: C=2.
y(1)=4π−1+eπ/42.
Let y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan−1x)dx=0,y(0)=1. Then the value of y(1) is :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
eπ/44+2π−1
eπ/42+4π−1
eπ/42−4π−1
eπ/44−2π−1
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