The solution of (1+x^2) frac{dy}{dx} + 2xy - | vedclass

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(B) Given the linear differential equation: $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$. Dividing by $(1+x^2)$,we get: $\frac{dy}{dx} + \left(\frac{2x}{1+x^2

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$3y(1+x^2) = 4x^3 + c$

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(B) Given the linear differential equation: $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$. Dividing by $(1+x^2)$,we get: $\frac{dy}{dx} + \left(\frac{2x}{1+x^2}\right)y = \frac{4x^2}{1+x^2}$. This is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1+x^2}$ and $Q = \frac{4x^2}{1+x^2}$. The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y(1+x^2) = \int \left(\frac{4x^2}{1+x^2}\right)(1+x^2) dx + c$. $y(1+x^2) = \int 4x^2 dx + c$. $y(1+x^2) = \frac{4x^3}{3} + c$. Multiplying by $3$,we get: $3y(1+x^2) = 4x^3 + c$.

The solution of $(1+x^2) \frac{dy}{dx} + 2xy - 4x^2 = 0$ is:

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