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

• A. $Ccos\left(kt\right)\lfloor C_1{e}^{\left(\sqrt{k/\alpha }\right)x}+C_2{e}^{-\left(\sqrt{k/\alpha }\right)x}\rfloor$
• B. $C{e}^{kt}\lfloor C_1{e}^{\left(\sqrt{k/\alpha }\right)x}+C_2{e}^{-\left(\sqrt{k/\alpha }\right)x}\rfloor$
• C. $C{e}^{kt}\lfloor C_{1}cos\left(\sqrt{k/\alpha }\right)x+C_{2}sin(-\sqrt{k/\alpha })x\rfloor$
• D. $Csinkt\lfloor C_{1}cos\left(\sqrt{k/\alpha }\right)x+C_{2}sin(-\sqrt{k/\alpha })x\rfloor$

Answer 1

The correct option is B $C{e}^{kt}\lfloor C_1{e}^{\left(\sqrt{k/\alpha }\right)x}+C_2{e}^{-\left(\sqrt{k/\alpha }\right)x}\rfloor$

$\frac{\partial u}{\partial t}=\alpha \frac{{\partial }^{2}u}{\partial x^2}$ ... (1)
Let $u=XT$ ....... (2)
is the solution of (1) then
$\frac{\partial u}{\partial t}=X.T^{\prime }$ and $\frac{{\partial }^{2}u}{\partial x^2}=X^{\ast }T$
Using separation of variable concept in (1)
$X{T}^{\prime }=\alpha X^{\ast }.T\Rightarrow \frac{\alpha X^{\ast }}{X}=\frac{T^{\prime }}{T}=K$
Taking $alpha{X}^{\ast }=KX$ we get
$(D^2-\frac{K}{\alpha })X=0\Rightarrow X=C_1{e}^{-\sqrt{k/\alpha }x}+C_2{e}^{\sqrt{k/\alpha }x}$
Taking $T^{\prime }=KT\Rightarrow \frac{T^{\prime }}{T}=K$
$\Rightarrow logT=kt+logC$
$\Rightarrow T=C{e}^{Kt}$
So by (2), solution is
$u=(C_1{e}^{-\sqrt{k/\alpha }x}+C_2{e}^{\sqrt{k/\alpha }x})C{e}^{Kt}$