A) \[c(y+1)({{e}^{x}}+1)+{{e}^{y}}=0\]
B) \[c(y+1)({{e}^{x}}-1)+{{e}^{y}}=0\]
C) \[c(y+1)({{e}^{x}}-1)-{{e}^{y}}=0\]
D) \[c(y+1)({{e}^{x}}+1)={{e}^{y}}\]
Correct Answer: D
Solution :
\[({{e}^{x}}+1)ydy=(y+1){{e}^{x}}dx\] Þ \[\left( \frac{y}{y+1} \right)dy=\left( \frac{{{e}^{x}}}{{{e}^{x}}+1} \right)\,dx\]Þ \[\left[ 1-\frac{1}{y+1} \right]dy=\left( \frac{{{e}^{x}}}{{{e}^{x}}+1} \right)\,dx\] Þ \[\int_{{}}^{{}}{\left\{ 1-\frac{1}{y+1} \right\}}dy=\int_{{}}^{{}}{\frac{{{e}^{x}}}{{{e}^{x}}+1}}dx\] Þ \[y=\log (y+1)+\log ({{e}^{x}}+1)+\log c\]or \[{{e}^{y}}=c(y+1)({{e}^{x}}+1)\].You need to login to perform this action.
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