A) \[xy=\frac{{{x}^{3}}}{3}+\frac{3}{2}{{x}^{2}}+2x+c\]
B) \[xy=\frac{{{x}^{4}}}{4}+{{x}^{3}}+{{x}^{2}}+c\]
C) \[xy=\frac{{{x}^{4}}}{4}+\frac{{{x}^{3}}}{3}+{{x}^{2}}+c\]
D) \[xy=\frac{{{x}^{4}}}{4}+{{x}^{3}}+{{x}^{2}}+cx\]
Correct Answer: A
Solution :
\[x\frac{dy}{dx}+y={{x}^{2}}+3x+2\] Þ \[\frac{dy}{dx}+\frac{y}{x}=x+3+\frac{2}{x}\] Here\[P=\frac{1}{x},\ Q=x+3+\frac{2}{x}\], therefore I.F. \[\frac{dy}{dx}=-1\] Now solve it.
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