- #1

Duderonimous

- 63

- 1

## Homework Statement

Let y[itex]_{1}[/itex](x,t)=Acos(k[itex]_{1}[/itex]x-ω[itex]_{1}[/itex]t) and y[itex]_{2}[/itex](x,t)=Acos(k[itex]_{2}[/itex]-ω[itex]_{2}[/itex]t) be two solutions to the wave equation

[itex]\frac{∂^{2}y}{∂x^{2}}[/itex]=[itex]\frac{1}{v^{2}}[/itex][itex]\frac{∂^{2}y}{∂t^{2}}[/itex]

for the same v. Show that y(x,t)=y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](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[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](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 ∂[itex]^{2}[/itex]y/∂t[itex]^{2}[/itex] over ∂[itex]^{2}[/itex]2/∂x[itex]^{2}[/itex] is equal to v[itex]^{2}[/itex].

I am hung up on the trig.

y(x,t)=y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t)

y(x,t)=Acos(k[itex]_{1}[/itex]x-ω[itex]_{1}[/itex]t)+Acos(k[itex]_{2}[/itex]-ω[itex]_{2}[/itex]t)

y(x,t)=A(cosk[itex]_{1}[/itex]xcosω[itex]_{1}[/itex]t+sink[itex]_{1}[/itex]xsinω[itex]_{1}[/itex]t)+A(cosk[itex]_{2}[/itex]xcosω[itex]_{2}[/itex]t+sink[itex]_{2}[/itex]xsinω[itex]_{2}[/itex]t)

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