Solution to the wave equation?

[Furbishkov]

Homework Statement

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)

Homework Equations

∂²y / ∂t² = v² (∂²y / ∂x²)

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

 

[Orodruin]

Homework Statement

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)

Homework Equations

∂²y / ∂t² = v² (∂²y / ∂x²)

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

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

 

[Furbishkov]

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

General form is : y(x,t) = Ae^i(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?

 

[Orodruin]

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