Answer to Question #87071 in Differential Equations for Pankaj

Question

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

Answer 1

The one-dimensional wave equation has the form

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

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

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

Then,

$$\frac{\partial u}{\partial x}=\frac{\partial}{\partial x}\left(\cos(\alpha cx)\cdot\sin(\alpha t)\right)=\frac{d}{dx}\left(\cos(\alpha cx)\right)\cdot\sin(\alpha t)=$$ $$=-c\alpha\cdot\sin(\alpha cx)\cdot\sin(\alpha t)\rightarrow$$ $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial x}\left(-c\alpha\cdot\sin(\alpha cx)\cdot\sin(\alpha t)\right)=$$ $$=\frac{d}{dx}\left(-c\alpha\cdot\sin(\alpha cx)\right)\cdot\sin(\alpha t)=-c^2\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)$$

Conclusion,

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

Then,

$$\frac{\partial u}{\partial t}=\frac{\partial}{\partial t}\left(\cos(\alpha cx)\cdot\sin(\alpha t)\right)=\cos(\alpha cx)\cdot\frac{d}{dt}\left(\sin(\alpha t)\right)=$$ $$=\alpha\cdot\cos(\alpha cx)\cdot\cos(\alpha t)\rightarrow$$ $$\frac{\partial^2 u}{\partial t^2}=\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial t}\right)=\frac{\partial}{\partial t}\left(\alpha\cdot\cos(\alpha cx)\cdot\cos(\alpha t)\right)=$$ $$=\cos(\alpha cx)\cdot\frac{d}{dt}\left(\alpha\cdot\cos(\alpha t)\right)=-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)$$

Conclusion,

$$\boxed{\frac{\partial^2 u}{\partial t^2}=-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)}$$

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

$$\frac{\partial^2 u}{\partial t^2}=c^2\cdot\frac{\partial^2 u}{\partial x^2}\rightarrow$$ $$-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=c^2\cdot\left(-c^2\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)\right)\rightarrow$$ $$-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=-c^4\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)\rightarrow$$ $$c^4\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)-\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=0\rightarrow$$ $$\left(c^4-1\right)\cdot\alpha^2\cdot\cos(\alpha cx)\cdot\sin(\alpha t)=0\rightarrow$$ $$c^4-1=0\rightarrow (c^2-1)(c^2+1)=0\rightarrow (c-1)(c+2)(c^2+1)=0\rightarrow$$ $$c-1=0\quad or\quad c+1=0\quad or\quad c^2+1=0\rightarrow$$ $$c=1\quad or\quad c=-1\quad or\quad c\in\varnothing$$

ANSWER

$$\boxed{c=\pm 1}$$