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

[Duderonimous]

Homework Statement

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.

Homework Equations

Equations given in problem statement and

cos(a-b)=cosacosb+sinasinb

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)

Cant figure out where to go from here. Please help! Thanks.

 

[vela]

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?

 

[Duderonimous]

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

 

[vela]

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
$$\frac{1}{v^2} \frac{\partial^2}{\partial t^2}y.$$

 

[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.