The solution of the differential equation (1 | vedclass

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(A) Given the differential equation: $(1 + x^2)\frac{dy}{dx} = x(1 + y^2)$Separate the variables $x$ and $y$:$\frac{1}{1 + y^2} dy = \frac{x}{1 + x^2}

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$2	an^{-1}y = \log(1 + x^2) + c$

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(A) Given the differential equation: $(1 + x^2)\frac{dy}{dx} = x(1 + y^2)$Separate the variables $x$ and $y$:$\frac{1}{1 + y^2} dy = \frac{x}{1 + x^2} dx$Integrate both sides:$\int \frac{1}{1 + y^2} dy = \int \frac{x}{1 + x^2} dx$Let $u = 1 + x^2$,then $du = 2x dx$,so $x dx = \frac{1}{2} du$:$	an^{-1}y = \frac{1}{2} \int \frac{1}{u} du$$	an^{-1}y = \frac{1}{2} \log(1 + x^2) + c_1$Multiply by $2$ to simplify:$2	an^{-1}y = \log(1 + x^2) + 2c_1$Let $c = 2c_1$:$2	an^{-1}y = \log(1 + x^2) + c$

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

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