The differential equation frac{dy}{dx} = frac | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{1}{ax + by + c}$.Taking the reciprocal of both sides,we get: $\frac{dx}{dy} = ax + by + c$

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linear in $x$

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{1}{ax + by + c}$.Taking the reciprocal of both sides,we get: $\frac{dx}{dy} = ax + by + c$.Rearranging the terms,we have: $\frac{dx}{dy} - ax = by + c$.This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -a$ and $Q = by + c$.Since the equation is in the form of a linear differential equation in $x$,the correct option is $B$.

The differential equation $\frac{dy}{dx} = \frac{1}{ax + by + c}$,where $a, b, c$ are all non-zero real numbers,is

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