# Solution to the wave equation?

1. Nov 11, 2015

### Furbishkov

1. The problem statement, all variables and given/known data
Is the function
y(x,t) = Ae(−β2x2−2βxt−t2) + Be(−x2+2αxt−α2t2)
a solution to the wave equation
2y / ∂t2 = v2 (∂2y / ∂x2)

2. Relevant equations

2y / ∂t2 = v2 (∂2y / ∂x2)

3. The attempt at a solution
I have found the solution through finding the partial derivatives (∂2y / ∂t2 and ∂2y / ∂x2) of the function. I got that it is a solution if α=-1 and β=1. However, this was a very long winded process of taking derivatives and am wondering if there is another way to solve it. I notice now that if I put α=-1 and β=1 into the original equation, y(x,t) = Ae(−β2x2−2βxt−t2) + Be(−x2+2αxt−α2t2) , then the exponents are the same. Am I suppose to notice this right away instead of taking the derivatives? Thanks

2. Nov 11, 2015

### Orodruin

Staff Emeritus
Yes, you are likely supposed to notice it directly. What is the general form of a solution to the wave equation?

3. Nov 12, 2015

### Furbishkov

General form is : y(x,t) = Aei(kx-ωt) = A[cos(kx-ωt) + isin(kx-ωt)]

So if I notice the exponents are the same under the condition α=-1 and β=1 how does that exactly translate to it being a solution of the wave equation?

4. Nov 12, 2015

### Orodruin

Staff Emeritus
No, this is not the most general form. It is just a plane wave solution.