Question #87071
Obtain the value of the constant c for which the function u(x,t) = cosαcxsinαt is a
solution of the wave equation . 2
2
2
2
2
x
1
Expert's answer
2019-03-27T11:11:03-0400

The one-dimensional wave equation has the form


2ut2=c22ux2\frac{\partial^2 u}{\partial t^2}=c^2\cdot\frac{\partial^2 u}{\partial x^2}

As you can see, you need to calculate the second partial derivatives of the specified function


u(x,t)=cos(αcx)sin(αt)u(x,t)=\cos(\alpha cx)\cdot\sin(\alpha t)

Then,


ux=x(cos(αcx)sin(αt))=ddx(cos(αcx))sin(αt)=\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\left(\cos(\alpha cx)\cdot\sin(\alpha t)\right)=\frac{d}{dx}\left(\cos(\alpha cx)\right)\cdot\sin(\alpha t)==cαsin(αcx)sin(αt)=-c\alpha\cdot\sin(\alpha cx)\cdot\sin(\alpha t)\rightarrow2ux2=x(ux)=x(cαsin(αcx)sin(αt))=\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial x}\left(-c\alpha\cdot\sin(\alpha cx)\cdot\sin(\alpha t)\right)==ddx(cαsin(αcx))sin(αt)=c2α2cos(αcx)sin(αt)=\frac{d}{dx}\left(-c\alpha\cdot\sin(\alpha cx)\right)\cdot\sin(\alpha t)=-c^2\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)

Conclusion,


2ux2=c2α2cos(αcx)sin(αt)\boxed{\frac{\partial^2 u}{\partial x^2}=-c^2\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)}

Then,


ut=t(cos(αcx)sin(αt))=cos(αcx)ddt(sin(αt))=\frac{\partial u}{\partial t}=\frac{\partial}{\partial t}\left(\cos(\alpha cx)\cdot\sin(\alpha t)\right)=\cos(\alpha cx)\cdot\frac{d}{dt}\left(\sin(\alpha t)\right)==αcos(αcx)cos(αt)=\alpha\cdot\cos(\alpha cx)\cdot\cos(\alpha t)\rightarrow2ut2=t(ut)=t(αcos(αcx)cos(αt))=\frac{\partial^2 u}{\partial t^2}=\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial t}\right)=\frac{\partial}{\partial t}\left(\alpha\cdot\cos(\alpha cx)\cdot\cos(\alpha t)\right)==cos(αcx)ddt(αcos(αt))=α2cos(αcx)sin(αt)=\cos(\alpha cx)\cdot\frac{d}{dt}\left(\alpha\cdot\cos(\alpha t)\right)=-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)

Conclusion,


2ut2=α2cos(αcx)sin(αt)\boxed{\frac{\partial^2 u}{\partial t^2}=-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)}


Substitute all found derivatives in the specified one-dimensional wave equation equation:


2ut2=c22ux2\frac{\partial^2 u}{\partial t^2}=c^2\cdot\frac{\partial^2 u}{\partial x^2}\rightarrowα2cos(αcx)sin(αt)=c2(c2α2cos(αcx)sin(αt))-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=c^2\cdot\left(-c^2\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)\right)\rightarrowα2cos(αcx)sin(αt)=c4α2cos(αcx)sin(αt)-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=-c^4\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)\rightarrowc4α2cos(αcx)sin(αt)α2cos(αcx)sin(αt)=0c^4\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=0\rightarrow(c41)α2cos(αcx)sin(αt)=0\left(c^4-1\right)\cdot\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=0\rightarrowc41=0(c21)(c2+1)=0(c1)(c+2)(c2+1)=0c^4-1=0\rightarrow (c^2-1)(c^2+1)=0\rightarrow (c-1)(c+2)(c^2+1)=0\rightarrowc1=0orc+1=0orc2+1=0c-1=0\quad or\quad c+1=0\quad or\quad c^2+1=0\rightarrowc=1orc=1orcc=1\quad or\quad c=-1\quad or\quad c\in\varnothing

ANSWER

c=±1\boxed{c=\pm 1}


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