Separate variables: 2xdx=1+ydy. Integrating, 21logx=log(1+y)+logc1, so logx=2log(1+y)+logc, giving x=(1+y)2c.
the general solution of the differential equation (1 + y) dx - 2x dy = 0 is :
CUET UG 2022 — Mathematics Applied-Mathematics
2022mcqmedium
the general solution of the differential equation (1+y)dx−2xdy=0 is :
Official previous-year question
Held on 11 Aug 2022 · Verified 13 Jul 2026.
Options
- A
x=(1+y)2c
- B
x2=(1+y)2c
- C
x2−y2=c
- D
y=c+x2y
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