xdxdy+y=ex⇒dxd(xy)=ex.
Integrating: xy=ex+C.
At (1,1): 1=e+C⇒C=1−e.
So xy=ex+1−e.
At x=−1: −y=e−1+1−e⇒y=e−1−e1=ee2−e−1=ee2−(e+1).
CUET UG 2022 — Mathematics Calculus
If curve represented by differential equation xdxdy+y=ex passes through (1, 1), then y(−1) is :
Held on 7 Aug 2022 · Verified 13 Jul 2026.
e
e−1
e2−(e+1)
ee2−(e+1)
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