A) \[4{{e}^{2x}}+{{e}^{-x}}+{{C}_{1}}x+{{C}_{2}}\]
B) \[\frac{1}{4}{{e}^{2x}}-{{e}^{-x}}+C\]
C) \[\frac{1}{4}{{e}^{2x}}+{{e}^{-x}}+{{C}_{1}}x+{{C}_{2}}\]
D) \[\frac{1}{4}{{e}^{2x}}-{{e}^{-x}}+{{C}_{1}}x+{{C}_{2}}\]
Correct Answer: C
Solution :
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{2x}}+{{e}^{-x}}\] ?..(i) Integrating on both sides w.r.t.x, \[\int{\frac{d}{dx}\left( \frac{dy}{dx} \right)}\,dx=\int{({{e}^{2x}}+{{e}^{-x}})dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{{{e}^{2x}}}{2}-{{e}^{-x}}+{{C}_{1}}\] Again integration on both sides w.r.t.x, \[\Rightarrow \] \[\int{\frac{d}{dx}\,(y)\,dx=\int{\left( \frac{{{e}^{2x}}}{2}-{{e}^{-x}}+{{C}_{1}} \right)}}dx\] \[\Rightarrow \] \[y=\frac{{{e}^{2x}}}{4}+{{e}^{-x}}+{{C}_{1}}x+{{C}_{2}}\]
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