```

```

# Solving Linear ODE

We now want to solve the ODEs that we got by simplifying the wave equation:

As you see, the two equations are very similar, except that one involves a function of time and the other involves a function of space, so we only need to solve one equation, and the other one will have the same solution. This type of linear ODE is usually solved by trigonomoteric substituion - assume that g(x) = Asin (kx) +Bcos(kx) and determine what k must be. Notice that we have two arbitrary constants A and B, which appear because we lose constants when we take derivatives, so in an equation with two derivatives there are always two arbitrary constants. We need additional conditions on the function in order to solve the equation completely, and we will do this in step 3.

Now that we have a guess, we need to show that it actually satisfies the equation. Taking the second derivative of Asin(kx) and Bcos(kx) results in the same function, only multiplied by -k^2. So we can write the equation now in terms of our sines and cosines:

-k^2(A sin(kx) + B cos (kx)) = -D(Asin(kx) + B cos (kx))

It is clear that in order for this equation to be true, k^2 must equal D. Now that we know k, we have the solution to this ODE (with some undetermined constants, still):

By essentially the same process, we determine that the solution to the time equation is:

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