Question

Consider the differential equation \(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 10{\rm{\;}}-{\rm{\;}}0.2{\rm{x}}\) with initial condition \({\rm{\;x}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1\). The response x(t) for \({\rm{t}} > 0\) is
1. \(2{\rm{\;}}-{\rm{\;}}{{\rm{e}}^{ - 0.2{\rm{t}}}}\)
2. \(2{\rm{\;}}-{\rm{\;}}{{\rm{e}}^{0.2{\rm{t}}}}\)
3. \(50{\rm{\;}}-{\rm{\;}}49{{\rm{e}}^{ - 0.2{\rm{t}}}}\)
4. \(50{\rm{\;}}-{\rm{\;}}49{{\rm{e}}^{0.2{\rm{t}}}}\)

Answer 1

Correct Answer - Option 3 : \(50{\rm{\;}}-{\rm{\;}}49{{\rm{e}}^{ - 0.2{\rm{t}}}}\)

\(\begin{array}{l} \frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 10 - 0.2{\rm{x}}\\ \frac{1}{{10 - \frac{{\rm{x}}}{5}}}.{\rm{dx}} = {\rm{dt}}\\ - 5\ln \left( {10 - \frac{{\rm{x}}}{5}} \right) = {\rm{t}} + {\rm{c}}\\ \left( {10 - \frac{{\rm{x}}}{5}} \right) = {{\rm{e}}^{ - {{\frac{1}{5}}^{\left( {{\rm{c}} + {\rm{t}}} \right)}}}}\\ \frac{{\rm{x}}}{5} = 10 - {{\rm{e}}^{ - \frac{{{\rm{c}} + {\rm{t}}}}{5}}}\\ {\rm{x}} = 50 - 5{{\rm{e}}^{ - \frac{{{\rm{c}} + {\rm{t}}}}{5}}} \end{array}\)

Given \({\rm{x}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1{\rm{\;\;\;}} \Rightarrow 1{\rm{\;}} = {\rm{\;}}50{\rm{\;}}-{\rm{\;}}5{{\rm{e}}^{ - \frac{{\rm{c}}}{5}{\rm{\;}}}}\)

\(\begin{array}{l} 5{{\rm{e}}^{ - \frac{{\rm{c}}}{5}}} = 49{\rm{\;}} \Rightarrow {{\rm{e}}^{ - \frac{{\rm{c}}}{5}}} = 9.8\\ \therefore {\rm{x}}\left( {\rm{t}} \right) = 50 - 5{{\rm{e}}^{ - \frac{{\rm{c}}}{5}}}.{\rm{\;}}{{\rm{e}}^{ - \frac{{\rm{t}}}{5}}} = 50 - 49{{\rm{e}}^{ - 0.2{\rm{t}}}} \end{array}\)