Question

The solution of the partial differential equation \(\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is of the form
1. \(C\cos \left( {kt} \right){C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}\)
2. \(C{e^{kt}}[{C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}]\)
3. \(C{e^{kt}}{C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x\)
4. \(C\sin \left( {kt} \right){C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x\)

Answer 1

Correct Answer - Option 2 : \(C{e^{kt}}[{C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}]\)

The PDE \(\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a one-dimensional heat equation

The solution of the one-dimensional heat equation is given by        

\({\rm{u}}\left( {{\rm{x}},{\rm{t}}} \right){\rm{\;}} = {\rm{\;}}\left( {{\rm{A\;cos\;px\;}} + {\rm{\;B\;sin\;px}}} \right){\rm{c}}{{\rm{e}}^{ - {{\rm{e}}^2}{{\rm{p}}^2}{\rm{t}}}}\)

Put –p²α = k

\(\Rightarrow p = \sqrt { - \frac{k}{\alpha }} = \sqrt {\frac{k}{\alpha }} i\)

Putting the value of p in eq. (i)

\(\begin{array}{l} u\left( {x,t} \right) = \left( {A\cos \sqrt {\frac{k}{\alpha }x + b} \sin h\;\sqrt {\frac{k}{\alpha }} x} \right)C{e^{kt}}\\ u\left( {x,t} \right) = C{e^{kt}}\left[ {A\left\{ {\frac{{{e^{\sqrt {\frac{k}{\alpha }} \;x}} + {e^{ - \sqrt {\frac{k}{\alpha }} }}x}}{2}} \right\} + B\left\{ {\frac{{{e^{\sqrt {\frac{k}{\alpha }} \;x}} - {e^{ - \sqrt {\frac{k}{\alpha }} }}x}}{2}} \right\}} \right]\\ u\left( {x,t} \right)= C{e^{kt}}\left[ {{e^{\sqrt {\frac{k}{\alpha }} \;x}}\left\{ {\frac{{A + B}}{2}} \right\} + {e^{ - \sqrt {\frac{k}{\alpha }} x}}\left\{ {\frac{{A - B}}{2}} \right\}} \right]\\ u\left( {x,t} \right)= C{e^{kt}}\;\left[ {{c_1}{e^{\sqrt {\frac{k}{\alpha }} x}} + {c_2}{e^{ - \sqrt {\frac{k}{\alpha }} \;x}}} \right] \end{array}\)