The differential equation 2xy , dy = (x^2 + y | vedclass

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(C) Given the differential equation: $2xy \, dy = (x^2 + y^2 + 1) dx$ Rearranging the terms: $2xy \, dy - y^2 dx = (x^2 + 1) dx$ Divide both sides by

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$A$ family of rectangular hyperbola with centre on $x$-axis

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(C) Given the differential equation: $2xy \, dy = (x^2 + y^2 + 1) dx$ Rearranging the terms: $2xy \, dy - y^2 dx = (x^2 + 1) dx$ Divide both sides by $x^2$: $\frac{2xy \, dy - y^2 dx}{x^2} = \frac{(x^2 + 1) dx}{x^2}$ This can be written as: $d\left(\frac{y^2}{x}\right) = (1 + x^{-2}) dx$ Integrating both sides: $\frac{y^2}{x} = x - \frac{1}{x} + C$ Multiplying by $x$: $y^2 = x^2 - 1 + Cx$ Rearranging to standard form: $x^2 - Cx - y^2 = 1$ Completing the square for $x$: $(x - \frac{C}{2})^2 - y^2 = 1 + \frac{C^2}{4}$ This equation represents a family of rectangular hyperbolas with the center at $(\frac{C}{2}, 0)$,which lies on the $x$-axis.

The differential equation $2xy \, dy = (x^2 + y^2 + 1) dx$ determines

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