A) \[\frac{2x}{{{x}^{2}}+1}\]
B) \[\frac{{{x}^{2}}-1}{{{x}^{2}}+1}\]
C) \[{{x}^{2}}+1\]
D) None of these
Correct Answer: C
Solution :
Given differential equation can be rewritten as \[\frac{dy}{dx}+\frac{2x}{1+{{x}^{2}}}y=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}\] Here, \[P=\frac{2x}{1+{{x}^{2}}}\]and\[Q=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}\] \[\therefore \] \[IF=\int{{{e}^{\frac{2x}{1+{{x}^{2}}}dx}}}={{e}^{\log (1+{{x}^{2}})}}\] \[=1+{{x}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
