A) \[3(1+{{x}^{2}})y=4{{x}^{3}}\]
B) \[3(1-{{x}^{2}})y=4{{x}^{3}}\]
C) \[3(1+{{x}^{2}})={{x}^{3}}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{dy}{dx}+\frac{2x}{1+{{x}^{2}}}y=\frac{4{{x}^{2}}}{1+{{x}^{2}}}\] It is linear equation of the form \[\frac{dy}{dx}+Py=Q\] Here \[P=\frac{2x}{1+{{x}^{2}}}\]and \[Q=\frac{4{{x}^{2}}}{1+{{x}^{2}}}\] I.F. \[={{e}^{\int_{{}}^{{}}{\frac{2x}{1+{{x}^{2}}}\,dx}}}={{e}^{\log (1+{{x}^{2}})}}=(1+{{x}^{2}})\] Therefore, solution is given by \[y.(1+{{x}^{2}})=\int{\frac{4{{x}^{2}}}{1+{{x}^{2}}}(1+{{x}^{2}})dx+c=\frac{4{{x}^{3}}}{3}}+c\]. But it passes through (0,0) therefore \[c=0\], hence the curve is \[3y(1+{{x}^{2}})=4{{x}^{3}}\].
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