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(C) Given the differential equation: $(x + 2y^3)\frac{dy}{dx} - y = 0$Rearranging the terms,we get: $(x + 2y^3)dy = y dx$$\Rightarrow \frac{dx}{dy} =

Vedclass · Accepted Answer

$x = y^3 + ky$

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(C) Given the differential equation: $(x + 2y^3)\frac{dy}{dx} - y = 0$Rearranging the terms,we get: $(x + 2y^3)dy = y dx$$\Rightarrow \frac{dx}{dy} = \frac{x + 2y^3}{y}$$\Rightarrow \frac{dx}{dy} = \frac{x}{y} + 2y^2$$\Rightarrow \frac{dx}{dy} - \frac{1}{y}x = 2y^2$This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = 2y^2$.The Integrating Factor ($I$.$F$.) is given by: $I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.The general solution is given by: $x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C$$\Rightarrow x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + C$$\Rightarrow \frac{x}{y} = \int 2y dy + C$$\Rightarrow \frac{x}{y} = y^2 + C$$\Rightarrow x = y^3 + Cy$Thus,the solution is $x = y^3 + Cy$.

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

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