A) \[y=x\text{ }log(x+1)-x-log(x+1)+5\]
B) \[y=x\text{ }log(x+1)+x+log(x+1)+5\]
C) \[y=x\text{ }log(x+1)-x+log(x+1)+5\]
D) None of the above
Correct Answer: C
Solution :
We have the differential equation, \[{{e}^{\frac{dy}{dx}}}=x+1\] Taking log on both sides, we get \[\frac{dy}{dx}=\log (x+1)\] \[\Rightarrow \] \[dy=\log (x+1)dx\] On integrating both sides, we get \[\int{dy=\int{\log (x+1)}}dx\] \[\Rightarrow \]\[y=\log (x+1)x-\int{\frac{1}{x+1}}.x\,dx+c\] \[\Rightarrow \]\[y=x\log (x+1)x-\int{\left( 1-\frac{1}{x+1} \right)}dx+c\] \[\Rightarrow \]\[y=x\log (x+1)-x+\log (x+1)+c\]...(i) But given that\[y(0)=5,\]so from Eq. (i), \[5=0-0+0+c\] \[\Rightarrow \] \[c=5\] Putting this value of c in Eq. (i), we get \[y=x\text{ }log(x+1)-x+log(x+1)+5\]which is the required solution.
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