A) \[4xy={{x}^{4}}+c\]
B) \[xy={{x}^{4}}+c\]
C) \[\frac{1}{4}xy={{x}^{4}}+c\]
D) \[xy=4{{x}^{4}}+c\]
Correct Answer: A
Solution :
The given equation \[\frac{dy}{dx}+\frac{y}{x}={{x}^{2}}\]is of the form \[\frac{dy}{dx}+Py=Q\]. So, I.F.= \[{{e}^{\int_{{}}^{{}}{\frac{1}{x}dx}}}={{e}^{\log x}}=x\] Hence required solution \[xy=\int_{{}}^{{}}{x.{{x}^{2}}dx+c}\] Þ \[xy=\frac{{{x}^{4}}}{4}+c\] Þ \[4xy={{x}^{4}}+c\].
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