A) \[y(1+{{x}^{3}})=x+\frac{1}{2}\sin 2x+c\]
B) \[y(1+{{x}^{3}})=cx+\frac{1}{2}\sin 2x\]
C) \[y(1+{{x}^{3}})=cx-\frac{1}{2}\sin 2x\]
D) \[y(1+{{x}^{3}})=\frac{x}{2}-\frac{1}{4}\sin 2x+c\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}+\frac{3{{x}^{2}}}{1+{{x}^{3}}}y=\frac{{{\sin }^{2}}x}{1+{{x}^{3}}}\] Here, \[P=\frac{3{{x}^{2}}}{1+{{x}^{3}}}\Rightarrow \]I.F. \[={{e}^{\int_{{}}^{{}}{P\,dx}}}={{e}^{\log (1+{{x}^{3}})}}=1+{{x}^{3}}\] Thus the solution is \[y.(1+{{x}^{3}})=\int_{{}}^{{}}{\frac{{{\sin }^{2}}x}{1+{{x}^{3}}}}(1+{{x}^{3}})dx=\int_{{}}^{{}}{\frac{1-\cos 2x}{2}}dx\] Þ \[y(1+{{x}^{3}})=\frac{1}{2}x-\frac{\sin 2x}{4}+c\].
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