The general solution of the differential equa | vedclass

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(A) Given the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$.Taking the reciprocal, we get $\frac{d x}{d y}=\frac{2 y^3-4 x y+y

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$4 x y^2+2 x=y^4+y^2+c$

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(A) Given the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$.Taking the reciprocal, we get $\frac{d x}{d y}=\frac{2 y^3-4 x y+y}{2 y^2+1} = \frac{y(2 y^2+1) - 4 x y}{2 y^2+1} = y - \frac{4 x y}{2 y^2+1}$.Rearranging the terms, we have $\frac{d x}{d y} + \left(\frac{4 y}{2 y^2+1}\right) x = y$.This is a linear differential equation of the form $\frac{d x}{d y} + P(y) x = Q(y)$, where $P(y) = \frac{4 y}{2 y^2+1}$ and $Q(y) = y$.The integrating factor $IF = e^{\int P(y) d y} = e^{\int \frac{4 y}{2 y^2+1} d y} = e^{\ln(2 y^2+1)} = 2 y^2+1$.The general solution is $x(IF) = \int Q(y)(IF) d y + C$.$x(2 y^2+1) = \int y(2 y^2+1) d y + C = \int (2 y^3+y) d y + C$.$x(2 y^2+1) = \frac{2 y^4}{4} + \frac{y^2}{2} + C = \frac{y^4}{2} + \frac{y^2}{2} + C$.Multiplying by $2$, we get $2 x(2 y^2+1) = y^4+y^2+2C$, which simplifies to $4 x y^2+2 x = y^4+y^2+C$.

The general solution of the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$ is

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