# Show that y(x,t)=y1(x,t)+y2(x,t) is a solution to the wave equation

1. Aug 28, 2012

### Duderonimous

1. The problem statement, all variables and given/known data
Let y$_{1}$(x,t)=Acos(k$_{1}$x-ω$_{1}$t) and y$_{2}$(x,t)=Acos(k$_{2}$-ω$_{2}$t) be two solutions to the wave equation
$\frac{∂^{2}y}{∂x^{2}}$=$\frac{1}{v^{2}}$$\frac{∂^{2}y}{∂t^{2}}$
for the same v. Show that y(x,t)=y$_{1}$(x,t)+y$_{2}$(x,t) is also a solution to the wave equation.

2. Relevant equations
Equations given in problem statement and

cos(a-b)=cosacosb+sinasinb

3. The attempt at a solution

I am pretty sure I put y$_{1}$(x,t)+y$_{2}$(x,t) in a form that does not involve addition like 2Acosasinb or something close to that. Then I take a second partial derivative with respect to both x and t and show that the ratio of ∂$^{2}$y/∂t$^{2}$ over ∂$^{2}$2/∂x$^{2}$ is equal to v$^{2}$.
I am hung up on the trig.

y(x,t)=y$_{1}$(x,t)+y$_{2}$(x,t)

y(x,t)=Acos(k$_{1}$x-ω$_{1}$t)+Acos(k$_{2}$-ω$_{2}$t)

y(x,t)=A(cosk$_{1}$xcosω$_{1}$t+sink$_{1}$xsinω$_{1}$t)+A(cosk$_{2}$xcosω$_{2}$t+sink$_{2}$xsinω$_{2}$t)

2. Aug 28, 2012

### vela

Staff Emeritus
You're making it too complicated. Don't use any trig identities. You know that y1 and y2 are solutions. What does that mean mathematically?

3. Aug 28, 2012

### Duderonimous

That their sum should also be a solution by the principle of superposition?

4. Aug 28, 2012

### vela

Staff Emeritus
Well, that's what you're supposed to show.

When you assume $y_i$ is a solution, you can say that
$$\frac{\partial^2}{\partial x^2}y_i(x,y) = \frac{1}{v^2}\frac{\partial^2}{\partial t^2} y_i(x,t),$$ where $\frac{\omega_i}{k_i} = v$.

So now plug $y=y_1+y_2$ into the lefthand side of the wave equation. You get
$$\frac{\partial^2}{\partial x^2}y = \frac{\partial^2}{\partial x^2}(y_1+y_2)$$ which you can split into two terms because differentiation is a linear operation. What you want to do is keep manipulating the righthand side until you end up with
$$\frac{1}{v^2} \frac{\partial^2}{\partial t^2}y.$$

5. Aug 29, 2012

### Duderonimous

$\frac{\partial^2}{\partial x^2}y = \frac{\partial^2}{\partial x^2}(y_1+y_2)$
$\frac{\partial^2}{\partial x^2}y = \frac{\partial^2 y_1}{\partial x^2} + \frac{\partial^2 y_2}{\partial x^2}$
$\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}\frac{\partial^2 y_1}{\partial t^2}+\frac{1}{v^2}\frac{\partial^2 y_1}{\partial t^2}$
$\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}(\frac{\partial^2 y_1}{\partial t^2}+\frac{\partial^2 y_2}{\partial t^2})$
$\frac{\partial^2}{\partial x^2}y= \frac{1}{v^2}\frac{\partial^2}{\partial t^2}(y_1+y_2)$
$\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}\frac{\partial^2}{\partial t^2}y$

Is that all I have to show since I already know that y is a solution to the wave equation and that it is assumed that is obeys that principle of superposition? Thanks again for your time.