A) \[y=\frac{{{\tan }^{-1}}x}{{{x}^{2}}+1}+C\]
B) \[y={{\tan }^{-1}}x+C\]
C) \[y({{x}^{2}}+1)={{\tan }^{-1}}x+C\]
D) \[y(ta{{n}^{-1}}x={{x}^{2}}+C\]
Correct Answer: C
Solution :
| \[{{({{x}^{2}}+1)}^{2}}\frac{dy}{dx}+2x({{x}^{2}}+1)y=1\] |
| \[\frac{dy}{dx}+\frac{2x}{{{x}^{2}}+1}y=\frac{1}{{{({{x}^{2}}+1)}^{2}}}\] |
| Linear D.E., |
| \[I.F.={{e}^{\int{\frac{2x}{{{x}^{2}}+1}dx}}}={{e}^{\ell n({{x}^{2}}+1)}}={{x}^{2}}+1\] |
| \[\therefore \] \[y(I.F.)=\int{Q(IF)dx}\] |
| \[y({{x}^{2}}+1)=\int{\frac{1}{{{({{x}^{2}}+1)}^{2}}}}({{x}^{2}}+1)dx\] |
| \[y({{x}^{2}}+1)={{\tan }^{-1}}x+C.\] |
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