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(B) Given the differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$ Divide by $(x^2+1)$ to get the standard linear form $\

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$y(x^2+1)^2 = x+c$

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(B) Given the differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$ Divide by $(x^2+1)$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} + \frac{4x}{x^2+1}y = \frac{1}{(x^2+1)^2}$ Here,$P(x) = \frac{4x}{x^2+1}$ and $Q(x) = \frac{1}{(x^2+1)^2}$. The Integrating Factor $(IF)$ is given by $e^{\int P(x) dx}$: $IF = e^{\int \frac{4x}{x^2+1} dx} = e^{2 \ln(x^2+1)} = e^{\ln(x^2+1)^2} = (x^2+1)^2$. The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$: $y(x^2+1)^2 = \int \left( \frac{1}{(x^2+1)^2} \cdot (x^2+1)^2 \right) dx + c$ $y(x^2+1)^2 = \int 1 dx + c$ $y(x^2+1)^2 = x + c$.

Solve the following differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$

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