Solve the following variable separable differential equation
dy/dx = (2 − y) / (x + 1)
dy/dx = (2 − y) / (x + 1)
clearly it is variable separable type
dy / (2 − y) = dx / (x + 1)
integrating both sides
∫ dy / (2 − y) = ∫ dx / (x + 1)
− ln|2 − y| = ln|x + 1| + lnC [ note the ( -1), coefficient of y when integrating]
0 = ln|2 − y| + ln|x + 1| + lnC
use the property of logarithms
0=ln [ |2 − y| |x + 1| C ]
or
[ |2 − y| |x + 1| C ] = 1
for more details refer to this video
cbse 12th applied mathematics 2025 2026 old board exam question papers applied mathematics
variable sepaprable differential equation
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