The solution of the equation (x + 2y^3)frac{d | vedclass

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(B) Given the equation $(x + 2y^3)\frac{dy}{dx} - y = 0$.Rearranging the terms,we get $(x + 2y^3)\frac{dy}{dx} = y$.This can be written as $\frac{dx}{

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$y^3 - x = Ay$

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(B) Given the equation $(x + 2y^3)\frac{dy}{dx} - y = 0$.Rearranging the terms,we get $(x + 2y^3)\frac{dy}{dx} = y$.This can be written as $\frac{dx}{dy} = \frac{x + 2y^3}{y}$.Simplifying,we get $\frac{dx}{dy} = \frac{x}{y} + 2y^2$,or $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$.This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = 2y^2$.The integrating factor $(I.F.)$ is given by $e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.The general solution is $x \cdot (I.F.) = \int Q \cdot (I.F.) dy + A$.Substituting the values,$x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + A$.$x \cdot \frac{1}{y} = \int 2y dy + A$.$x \cdot \frac{1}{y} = y^2 + A$.Multiplying by $y$,we get $x = y^3 + Ay$,which can be rewritten as $y^3 - x = -Ay$. Since $A$ is an arbitrary constant,we can write the solution as $y^3 - x = Ay$.

The solution of the equation $(x + 2y^3)\frac{dy}{dx} - y = 0$ is

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