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Question

(B) Given the differential equation $\frac{dy}{dx} = \frac{y}{y^2 - x}$.Taking the reciprocal,we get $\frac{dx}{dy} = \frac{y^2 - x}{y} = y - \frac{x}

Vedclass · Accepted Answer

$y^3 - 3xy = c$

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(B) Given the differential equation $\frac{dy}{dx} = \frac{y}{y^2 - x}$.Taking the reciprocal,we get $\frac{dx}{dy} = \frac{y^2 - x}{y} = y - \frac{x}{y}$.Rearranging the terms,we obtain the linear differential equation in $x$: $\frac{dx}{dy} + \frac{1}{y}x = y$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y}$ and $Q(y) = y$.The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int \frac{1}{y} dy} = e^{\ln|y|} = y$.The general solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + c$.Substituting the values,$x \cdot y = \int y \cdot y dy + c$.$xy = \int y^2 dy + c$.$xy = \frac{y^3}{3} + c$.Multiplying by $3$,we get $3xy = y^3 + 3c$,which can be written as $y^3 - 3xy = c'$ (where $c' = -3c$).Thus,the correct option is $B$.

The solution of the differential equation $\frac{dy}{dx} = \frac{y}{y^2 - x}$ is

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