A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: A
Solution :
[a] \[{{e}^{x+y}}+{{e}^{y-x}}=1\] \[\Rightarrow \,\,\,\,\,\,\,({{e}^{x}}+{{e}^{-x}})={{e}^{-y}}\] ...(1) Differentiating both sides w.r.t. x, we get \[{{e}^{x}}-{{e}^{-x}}=-{{e}^{-y}}\frac{dy}{dx}\] ?..(2) Again, differentiating both sides w.r.t. x, we get \[-({{e}^{x}}+{{e}^{-x}})={{e}^{-y}}.\frac{{{d}^{2}}y}{d{{x}^{2}}}-{{e}^{-y}}{{\left( \frac{dy}{dx} \right)}^{2}}\] \[\therefore \,\,\,\,\,y''-{{(y')}^{2}}+1=0\] [Using (1)] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,c=1\]
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