Question

The solution of the differential equation \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0\) with \({\rm{y}}\left( 0 \right) = {\rm{\;y'}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1{\rm{\;}}\)is
1. \(\left( {2{\rm{\;}} - {\rm{\;t}}} \right){{\rm{e}}^{\rm{t}}}\)
2. \(\left( {1{\rm{\;}} + {\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{ - {\rm{t}}}}\)
3. \(\left( {2{\rm{\;}} + {\rm{\;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}\)
4. \(\left( {1{\rm{\;}}-{\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{\rm{t}}}\)

Answer 1

Correct Answer - Option 2 : \(\left( {1{\rm{\;}} + {\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{ - {\rm{t}}}}\)

\(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2.\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0\)

Characteristics solution \({{\rm{Y}}_{\rm{c}}}{\rm{\;}} = {\rm{\;}}\left( {{{\rm{C}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{C}}_2}{\rm{\;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}{\rm{\;}}\)

\(\begin{array}{l} {\rm{Y\;}}\left( 0 \right){\rm{\;}} = {\rm{\;}}{{\rm{C}}_1}{\rm{\;}} = {\rm{\;}}1\\ {{\rm{Y}}_1}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}} - {\rm{\;}}\left( {{{\rm{C}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{C}}_2}{\rm{\;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{ - {\rm{t}}}}.{\rm{\;}}{{\rm{C}}_2}\\ {{\rm{Y}}_1}{\rm{\;}}\left( 0 \right){\rm{\;}} = {\rm{\;}} - {\rm{\;}}{{\rm{C}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{C}}_2}{\rm{\;}} = {\rm{\;}}1{\rm{\;\;\;}} \Rightarrow {\rm{\;}}{{\rm{C}}_2}{\rm{\;}} = {\rm{\;}}2\\ \therefore {\rm{\;y\;}} = {\rm{\;}}\left( {1{\rm{\;}} + {\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{ - {\rm{t}}}} \end{array}\)