A) \[{{x}^{3}}y+\frac{{{x}^{4}}}{4}+c=0\]
B) \[{{x}^{3}}y=\frac{{{x}^{4}}}{4}+c\]
C) \[{{x}^{3}}y+\frac{{{x}^{4}}}{4}=0\]
D) None of these
Correct Answer: B
Solution :
\[x\frac{dy}{dx}+3y=x\] Þ \[\frac{dy}{dx}+\frac{3y}{x}=1\] It is in the form of \[\frac{dy}{dx}+Py=Q\] So, I.F. \[={{e}^{\int_{{}}^{{}}{Pdx}}}={{e}^{3}}^{\int_{{}}^{{}}{\frac{1}{x}dx}}={{e}^{3\log x}}={{x}^{3}}\] Hence required solution is \[y+{{x}^{2}}+2x+2=c{{e}^{x}}\] Þ \[y{{x}^{3}}=\frac{{{x}^{4}}}{4}+c\].
You need to login to perform this action.
You will be redirected in
3 sec
