A) \[y=({{c}_{1}}+{{c}_{2}}x){{e}^{x}}+\frac{{{e}^{3x}}}{8}\]
B) \[y=({{c}_{1}}+{{c}_{2}}x){{e}^{-x}}+\frac{{{e}^{-3x}}}{8}\]
C) \[y=({{c}_{1}}+{{c}_{2}}x){{e}^{-x}}+\frac{{{e}^{3x}}}{8}\]
D) \[y=({{c}_{1}}+{{c}_{2}}x){{e}^{x}}+\frac{{{e}^{-3x}}}{8}\]
Correct Answer: C
Solution :
The given equation can be written as \[({{D}^{2}}+2D+1)t=2{{e}^{3x}},\]where \[\frac{d}{dx}=D\] Here, \[F(D)={{D}^{2}}+2D+1\] and\[Q=2{{e}^{3x}}\] The auxiliary equation is \[{{m}^{2}}+2m+1=0\] \[\Rightarrow \] \[{{(m+1)}^{2}}=0\] \[\Rightarrow \] \[m=-1,\,-1\] \[\therefore \] The CF \[=({{c}_{1}}+{{c}_{2}}x){{e}^{-x}}\] and \[PI=\frac{1}{F(D)}2{{e}^{3x}}=2\cdot \frac{1}{{{D}^{2}}+2D+1}\cdot {{e}^{3x}}\] \[=2\cdot \frac{{{e}^{3x}}}{9+6+1}=\frac{{{e}^{3x}}}{8}\] Hence, the complete solution is \[y=CF+PI\] \[\Rightarrow \] \[y={{({{c}_{1}}+{{c}_{2}}x)}^{{{e}^{-x}}}}+\frac{{{e}^{3x}}}{8}.\]
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