The solution of the D.Exx2+1dydx=y1−x2+x3log⁡x is

dydx+x2−1xx2+1y=x2log⁡xx2+1I.F.=e∫x2−1xx2+1dx=e∫2xx2+1−1x dx=elog⁡x2+1x=x2+1x Solution of the D.E yx2+1x=∫xlog⁡xdxyx2+1x=logx ∫xdx -∫1x∫xdx dxyx2+1x=x22log⁡x-∫1x·x22dxyx2+1x=x22log⁡x-x24+c

The solution of the D.Exx2+1dydx=y1−x2+x3log⁡x is

dydx+x2−1xx2+1y=x2log⁡xx2+1I.F.=e∫x2−1xx2+1dx=e∫2xx2+1−1x dx=elog⁡x2+1x=x2+1x Solution of the D.E yx2+1x=∫xlog⁡xdxyx2+1x=logx ∫xdx -∫1x∫xdx dxyx2+1x=x22log⁡x-∫1x·x22dxyx2+1x=x22log⁡x-x24+c

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$The solution of the D . E x (x^{2} + 1) \frac{d y}{d x} = y (1 - x^{2}) + x^{3} \log x is$

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$\frac{{y(x}^{2} -1)}{x} = \frac{1}{2} log (x) {+x}^{2} / 2 +C$

$\frac{y^{2} {(x}^{2} -1)}{x} = \frac{1}{2} x^{2} logx- \frac{1}{4} x^{2} +C$

$\frac{{y(x}^{2} +1)}{x} = \frac{1}{4} x^{2} logx+ \frac{x^{2}}{2} +C$

$\frac{{y(x}^{2} +1)}{x} = \frac{1}{2} x^{2} logx- \frac{x^{2}}{4} +C$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{dy}{dx} + \frac{x^{2} - 1}{x (x^{2} + 1)} y = \frac{x^{2} \log x}{x^{2} + 1} \\ I . F . = e^{\int \frac{x^{2} - 1}{x (x^{2} + 1)} d x} \\ = e^{\int (\frac{2 x}{x^{2} + 1} - \frac{1}{x}) d x} \\ = e^{\log \frac{(x^{2} + 1)}{x}} \\ = \frac{(x^{2} + 1)}{x} \\ Solution of the D.E \frac{y (x^{2} + 1)}{x} = \int xlog xdx \\ \frac{y (x^{2} + 1)}{x} = \log x \int x d x - \int (\frac{1}{x} \int x d x) d x \\ \frac{y (x^{2} + 1)}{x} = \frac{x^{2}}{2} \log x - \int \frac{1}{x} \cdot \frac{x^{2}}{2} dx \\ \frac{y (x^{2} + 1)}{x} = \frac{x^{2}}{2} \log x - \frac{x^{2}}{4} + c \end{array}$