A) \[-1\]
B) \[0\]
C) \[1\]
D) \[e\]
Correct Answer: B
Solution :
\[y'(x)+y(x)g'(x)=g(x)g'(x),y(0)=0;\] \[\forall \,x\in R\] \[\frac{d}{dx}\left( y\left( x \right) \right)+y\left( x \right)g'\left( x \right)=g\left( x \right)g'\left( x \right)\] \[g(0)=g(2)=0\] I.f. \[={{e}^{\int{g'(x)\,dx}}}={{e}^{g(x)}}\] \[y\left( x \right){{e}^{g(x)}}=\int{{{e}^{g\left( x \right)}}g\left( x \right)g'\left( x \right)dx+c}\] Let \[g\left( x \right)=t\Rightarrow g'\left( x \right)dx=dt\] \[y\left( x \right){{e}^{g\left( x \right)}}=\int{t{{e}^{t}}dt=t{{e}^{t}}-{{e}^{t}}+c}\] \[y\left( x \right)=\left( g\left( x \right)-1 \right)+c{{e}^{-g\left( x \right)}}\] Let \[x=0,\,y\left( 0 \right)=\left( g\left( 0 \right)-1 \right)+c{{e}^{-g\left( 0 \right)}}\] \[\Rightarrow \,\,0=\left( 0-1 \right)+c\Rightarrow c=1\] \[y\left( x \right)=\left( g\left( x \right)-1 \right)+{{e}^{-g\left( x \right)}}\] \[\Rightarrow y\left( 2 \right)=\left( g\left( 2 \right)-1 \right)+{{e}^{-g\left( 2 \right)}}\] \[\Rightarrow \,y\left( 2 \right)=\left( 0-1 \right)+{{e}^{-\left( 0 \right)}}=-1+1=0\]
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