A) \[{{e}^{y}}(x+1)=x+C\]
B) \[{{e}^{-y}}=2x+C\]
C) \[{{e}^{y}}(x+1)=2x+C\]
D) \[{{e}^{y}}(x+1)=C\]
Correct Answer: C
Solution :
[c] \[(x+1)\frac{dy}{dx}+1=2{{e}^{-y}}\] \[\Rightarrow {{e}^{y}}\frac{dy}{dx}+\frac{{{e}^{y}}}{x+1}=\frac{2}{x+1}\] Put \[{{e}^{y}}=u\Rightarrow {{e}^{y}}\frac{dy}{dx}=\frac{du}{dx}\,\,\,\,\,\therefore \,\,\frac{du}{dx}+\frac{u}{x+1}=\frac{2}{x+1}\]I.F. \[={{e}^{\int{\frac{1}{x+1}dx}}}={{e}^{\log (x+1)}}=x+1\] \[\therefore \] Solution is \[u(x+1)=\int{\frac{2}{(x+1)}(x+1)dx+C}\] \[\Rightarrow {{e}^{y}}(x+1)=2x+C\]
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