Question

$\footnotesize\textrm{Show that if a is a constant, then u(x,t)=sin(at)cos(x) is a solution to \dfrac{\partial^2 u}{\partial t^2}=a^2 \dfrac{\partial^2 u}{\partial x^2} }$

Answer 1

Refer to the above both attachments :D

Used Concepts :-

• ${\bf \dfrac{d}{dx} \{C \cdot f ( x )\} = C \cdot \dfrac{d}{dx} \{f(x)\}}$

• ${\bf \dfrac{d}{dx} \{\sin ( x)\}= \cos ( x)}$

• ${\bf \dfrac{d}{dx} \{\cos ( x)\}= - \sin ( x)}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: u(x,t)=sin(at) \: cos(x)$

that can be further rewritten as

$\rm \: u=sin(at) \: cos(x) - - - (1)$

Differentiate partially w. r. t. x, we get

$\rm \: \dfrac{\partial }{\partial x}u=\dfrac{\partial }{\partial x} \: sin(at) \: cos(x)$

$\rm \: \dfrac{\partial u}{\partial x}=sin(at) \: \dfrac{\partial }{\partial x} \: cos(x)$

$\rm \: \dfrac{\partial u}{\partial x}=sin(at) \: ( - sinx)$

$\rm \: \dfrac{\partial u}{\partial x} \: = - \: sin(at) \:sinx$

On differentiating partially w. r. t. x, we get

$\rm \:\dfrac{\partial }{\partial x} \dfrac{\partial u}{\partial x} \: = -\dfrac{\partial }{\partial x} \: sin(at) \:sinx$

$\rm \:\dfrac{\partial^2 u}{\partial x^2} \: = - \: sin(at)\dfrac{\partial }{\partial x} \:sinx$

$\rm\implies \:\rm \:\dfrac{\partial^2 u}{\partial x^2} \: = - \: sin(at) \: cosx - - - - (2)$

Now, Consider

$\rm \: u=sin(at) \: cos(x)$

On differentiating partially w. r. t. t, we get

$\rm \:\dfrac{\partial }{\partial t} u=\dfrac{\partial }{\partial t}sin(at) \: cos(x)$

$\rm \:\dfrac{\partial u}{\partial t} =cosx \: \dfrac{\partial }{\partial t}sin(at) \:$

$\rm \:\dfrac{\partial u}{\partial t} =cosx \: cos(at) \: a\:$

$\rm \:\dfrac{\partial u}{\partial t} =a \: cosx \: cos(at) \:$

On differentiating both sides w. r. t. t, we get

$\rm \:\dfrac{\partial }{\partial t}\dfrac{\partial u}{\partial t} =\dfrac{\partial }{\partial t}a \: cosx \: cos(at) \:$

$\rm \:\dfrac{\partial^2 u}{\partial t^2} = \: a \: cosx \: \dfrac{\partial }{\partial t} \: cos(at) \:$

$\rm \:\dfrac{\partial^2 u}{\partial t^2} = \: a \: cosx \:( - \: sin \: at) \: a \:$

$\rm \:\dfrac{\partial^2 u}{\partial t^2} = \: - \: {a}^{2} \: cosx \:\: sin(at) \:$

can be further rewritten as

$\rm \:\dfrac{\partial^2 u}{\partial t^2} = \: {a}^{2} \: [ - \: cosx \:\: sin(at)] \:$

So, using equation (1), can be further rewritten as

$\rm \:\dfrac{\partial^2 u}{\partial t^2} = \: {a}^{2} \: \dfrac{\partial^2 u}{\partial x^2} \:$

Hence,

$\footnotesize\textrm{If a is a constant, then u(x,t)=sin(at)cos(x) is a solution to \dfrac{\partial^2 u}{\partial t^2}=a^2 \dfrac{\partial^2 u}{\partial x^2} }$

$\rule{190pt}{2pt}$

Formulae used :-

$\boxed{\rm{  \:\dfrac{d}{dx} \: sinx \: = \: cosx \: }}$

$\boxed{\rm{  \:\dfrac{d}{dx} \: cosx \: = \: - \: sinx \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$