A) \[y=x\log x-x+2\]
B) \[y=(x+1)log|x+1|-x+3\]
C) \[y=(x+1)log|x+1|+x+3\]
D) \[y=xlogx+x+3\]
Correct Answer: B
Solution :
[b]: We have, \[{{e}^{\frac{dy}{dx}}}=x+1\] \[\Rightarrow \]\[\frac{dy}{dx}={{\log }_{e}}(x+1)\] \[\Rightarrow \]\[\int_{{}}^{{}}{dy}=\int_{{}}^{{}}{{{\log }_{e}}}(x+1)dx\] \[\Rightarrow \]\[y=(x+1)log|x+1|-(x+1)+c\] When \[y(0)=3\] then, \[3=0-1+c\Rightarrow c=4\] \[\Rightarrow \]\[y=(x+1)log\left| x+1 \right|-x+3\]You need to login to perform this action.
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