A) \[\log x\]
B) \[x\]
C) \[{{e}^{x}}\]
D) \[\frac{1}{x}\]
E) \[\frac{-1}{x}\]
Correct Answer: B
Solution :
\[dy(x+{{x}^{3}})=-dx(1+y+{{x}^{2}}y)\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{1+y+{{x}^{2}}y}{x+{{x}^{3}}}\] \[\Rightarrow \] \[\frac{dy}{dx}+\frac{y{{(1+x)}^{2}}}{x(1+{{x}^{2}})}=-\frac{1}{x(1+{{x}^{2}})}\] \[\Rightarrow \] \[\frac{dy}{dx}+\frac{y}{x}=-\frac{1}{x(1+{{x}^{2}})}\] \[\therefore \] \[IF={{e}^{\int{\frac{1}{x}dx}}}={{e}^{\log x}}=x\]
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