A) \[y=\frac{{{x}^{4}}}{5}+Cx\]
B) \[y=\frac{{{x}^{3}}}{3}+Cx\]
C) \[y=\frac{{{x}^{3}}}{3}+C\]
D) \[y=\frac{{{x}^{4}}}{4}+x\]
E) None of these
Correct Answer: E
Solution :
We have, \[\frac{dy}{dx}+\frac{y}{x}={{x}^{3}}\] On comparing with \[\frac{dy}{dx}+py=Q,\] we get \[P=\frac{1}{x},Q={{x}^{3}}\] \[\therefore \] \[IF={{e}^{\int{\frac{1}{x}dx}}}={{e}^{{{\log }_{e}}x}}=x\] Now, solution of given differential equation is \[y.If=\int{!Q.\,IF\,dx+C}\] \[\Rightarrow \] \[yx=\int{{{x}^{3}}.x\,dx+C}\] \[\Rightarrow \] \[yx=\int{{{x}^{4}}\,dx+C}\] \[\Rightarrow \] \[yx=\frac{{{x}^{5}}}{5}+C\] \[\Rightarrow \] \[y=\frac{{{x}^{4}}}{5}+\frac{C}{x}\]
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