A) \[y(1+{{x}^{2}})=c+{{\tan }^{-1}}x\]
B) \[\frac{y}{1+{{x}^{2}}}=c+{{\tan }^{-1}}x\]
C) \[y\log (1+{{x}^{2}})=c+{{\tan }^{-1}}x\]
D) \[y(1+{{x}^{2}})=c+{{\sin }^{-1}}x\]
Correct Answer: A
Solution :
| [a] Given differential equation is |
| \[\frac{dy}{dx}+\frac{2yx}{1+{{x}^{2}}}=\frac{1}{{{(1+{{x}^{2}})}^{2}}}\] which is a linear. |
| Differential equation of the form: \[\frac{dy}{dx}+Py=Q\] |
| On comparing, we have |
| \[P=\frac{2x}{1+{{x}^{2}}}\] and \[Q=\frac{1}{{{(1+{{x}^{2}})}^{2}}}\] |
| I.F \[=\,\,{{e}^{\int{2x/1+{{x}^{2}}dx}}}={{e}^{\log (1+{{x}^{2}})}}\,=\,\,(1+{{x}^{2}})\] |
| \[\therefore \] Solutions \[y(1+{{x}^{2}})\] |
| \[=\int{\frac{1}{{{(1+{{x}^{2}})}^{2}}}(1+{{x}^{2}})dx+c}\] |
| \[\Rightarrow y(1+{{x}^{2}})=\int{\frac{1}{(1+{{x}^{2}})}dx+c}\] |
| \[\Rightarrow y(1+{{x}^{2}})={{\tan }^{-1}}x+c\] |
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