The solution of the equation (x-4y^3) frac{dy | vedclass

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(B) Given the differential equation: $(x-4y^3) \frac{dy}{dx}-y=0$ for $y>0$. Rearranging the terms,we get: $(x-4y^3) \frac{dy}{dx}=y$. Since $y>0$,we

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$x+2y^3=cy$

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(B) Given the differential equation: $(x-4y^3) \frac{dy}{dx}-y=0$ for $y>0$. Rearranging the terms,we get: $(x-4y^3) \frac{dy}{dx}=y$. Since $y>0$,we can write: $\frac{dx}{dy} = \frac{x-4y^3}{y} = \frac{x}{y} - 4y^2$. This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -\frac{1}{y}$ and $Q = -4y^2$. The integrating factor ($I$.$F$.) is given by: $I.F. = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = e^{\ln y^{-1}} = \frac{1}{y}$. The general solution is: $x \cdot (I.F.) = \int (Q \cdot I.F.) dy + C$. Substituting the values: $x \cdot \frac{1}{y} = \int (-4y^2 \cdot \frac{1}{y}) dy + C$. $\frac{x}{y} = \int -4y dy + C$. $\frac{x}{y} = -4 \cdot \frac{y^2}{2} + C = -2y^2 + C$. Multiplying by $y$,we get: $x = -2y^3 + Cy$,which simplifies to $x+2y^3=Cy$.

The solution of the equation $(x-4y^3) \frac{dy}{dx}-y=0, (y>0)$ is

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