A) \[3x(1+{{y}^{2}})=4{{y}^{3}}+c\]
B) \[3y(1+{{x}^{2}})=4{{x}^{3}}+c\]
C) \[3x(1-{{y}^{2}})=4{{y}^{3}}+c\]
D) \[3y(1+{{y}^{2}})=4{{x}^{3}}+c\]
Correct Answer: B
Solution :
Given differential equation can be rewritten as \[\frac{dy}{dx}+\frac{2x}{1+{{x}^{2}}}y=\frac{4{{x}^{2}}}{1+{{x}^{2}}}\] It is a linear differential equation of the form \[\frac{dy}{dx}+Py=Q\] \[\therefore \] \[={{e}^{\int_{{}}^{{}}{\frac{2x}{1+{{x}^{2}}}dx}}}=(1+{{x}^{2}})\] \[\therefore \] Solution is \[y.(1+{{x}^{2}})=\int_{{}}^{{}}{(1+{{x}^{2}}).\frac{4{{x}^{2}}}{1+{{x}^{2}}}}dx\] \[\Rightarrow \]\[y(1+{{x}^{2}})=\frac{4{{x}^{3}}}{3}+{{c}_{1}}\] \[\Rightarrow \]\[3y(1+{{x}^{2}})=4{{x}^{3}}+c\]
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