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CUET
-- Mathematics - Section B1
Differential Equations
a
b
c
d
so $C=-ln|50|$
and $-ln|50-y|=kx-ln|50|$
$ln|50-y|-ln|50|=-kx$
so $ln\left|1-\frac{y}{50}\right|=-kx$
$1-\frac{y}{50}=e^{-kx}$
so $y=50(1-e^{-kx})$
