A) \[y=(x+1){{e}^{3x}}+c\]
B) \[3y=(x+1)+{{e}^{3x}}+c\]
C) \[\frac{3y}{x+1}={{e}^{3x}}+c\]
D) \[y{{e}^{-3x}}=3(x+1)+c\]
Correct Answer: C
Solution :
[c] The given equation is \[\frac{dy}{dx}-\frac{y}{x+1}={{e}^{3x}}(x+1)\] I.F. \[={{e}^{\int{\frac{1}{x+1}dx}}}={{e}^{-\log (x+1)}}=\frac{1}{x+1}\] The solution is \[y\left( \frac{1}{x+1} \right)=\int{{{e}^{3x}}(x+1).\frac{1}{x+1}}dx+a\] \[\Rightarrow \frac{y}{x+1}=\int{{{e}^{3x}}dx+a=\frac{{{e}^{3x}}}{3}+a}\] \[\Rightarrow \frac{3y}{x+1}={{e}^{3x}}+c,c=3a\]
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