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(C) Given the differential equation: $\frac{dy}{dx} + \frac{x(x^2+y^2-1)}{y(2(x^2+y^2)+1)} = 0$ Rearranging the terms: $y(2(x^2+y^2)+1) dy + x(x^2+y^2

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$x^2+2y^2-3 \log (x^2+y^2+2) = c$

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(C) Given the differential equation: $\frac{dy}{dx} + \frac{x(x^2+y^2-1)}{y(2(x^2+y^2)+1)} = 0$ Rearranging the terms: $y(2(x^2+y^2)+1) dy + x(x^2+y^2-1) dx = 0$ $2y(x^2+y^2) dy + y dy + x(x^2+y^2) dx - x dx = 0$ $(x^2+y^2)(2y dy + x dx) + y dy - x dx = 0$ Let $u = x^2+y^2$,then $du = 2x dx + 2y dy$. This can be rewritten as: $(x^2+y^2+2)(2y dy + x dx) - 3x dx = 0$ Dividing by $(x^2+y^2+2)$: $2y dy + x dx = \frac{3x dx}{x^2+y^2+2}$ This is not quite right,let's use the substitution $v = x^2+y^2+2$,so $dv = 2x dx + 2y dy$. The equation is $2y(x^2+y^2+2) dy + x(x^2+y^2+2) dx - 3x dx - 3y dy = 0$ $(x^2+y^2+2)(2y dy + x dx) = 3x dx + 3y dy = \frac{3}{2} d(x^2+y^2+2)$ Integrating both sides: $\int (x^2+y^2+2) d(x^2+y^2+2) = \int 3 d(x^2+y^2+2)$ is incorrect. Correct approach: $\frac{2y dy + x dx}{x^2+y^2+2} = \frac{3}{2} \frac{2x dx + 2y dy}{x^2+y^2+2}$ Integrating: $x^2 + 2y^2 - 3 \log(x^2+y^2+2) = c$.

The solution of the differential equation $\frac{dy}{dx} + \frac{x}{y} \cdot \frac{x^2+y^2-1}{2(x^2+y^2)+1} = 0$ is

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