Converting Cos and Sin to Exponential in Differential Equations

[jesuslovesu]

Homework Statement

I am in the process of doing a physics problem with a differential equation that has the form:

y = Acos(kx) + Bsin(kx)

According to my notes, this can also be written as y =Ae^jkx + Be^-jkx, unfortunately I just don't see how to write the original equation like that.

Homework Equations

The Attempt at a Solution

I know that cos(x) = 1/2[ e^jx + e^-jx ]
sin(x) = 1/(2j) [ e^jx - e^-jx ]

I can almost see how you would get it for the cos(kx) term:
Since Real { cos(kx) + j sin(kx) } = e^jkx using Euler's identity.
But for sine, I am stumped.

 

[Ben Niehoff]

The A and B in the first equation are not the same as the A and B in the second equation. Give them all different letters. Then you can find the coefficients in the second equation in terms of the coefficients in the first (by means of a system of 2 linear equations).

 

[Architeuthis Dux]

I can provide a response to this content by explaining the process of converting cosine and sine functions to exponential form in differential equations.

Firstly, it is important to understand that the exponential form of a complex number is given by ejx = cos(x) + jsin(x), where j is the imaginary unit. This is known as Euler's identity and is a fundamental concept in mathematics.

Now, let's look at the given equation y = Acos(kx) + Bsin(kx). We can rewrite this as y = A(cos(kx) + jsin(kx)) + B(sin(kx) + jcos(kx)). Using Euler's identity, we can further simplify this to y = Ae^jkx + Be^-jkx.

To understand how this conversion works, let's focus on the cos(kx) term. As mentioned earlier, ejkx = cos(kx) + jsin(kx). Therefore, we can rewrite cos(kx) as 1/2(ejkx + e^-jkx). Similarly, for the sin(kx) term, we can write it as 1/(2j)(ejkx - e^-jkx).

Therefore, the final form of the equation becomes y = 1/2(A + Bj)(ejkx + e^-jkx) + 1/2j(A - Bj)(ejkx - e^-jkx). By simplifying the terms, we get y = Aejkx + Be^-jkx, which is the desired exponential form.

In summary, converting cosine and sine functions to exponential form in differential equations involves using Euler's identity and simplifying the terms using basic algebraic operations. With practice, this process can become easier and help in solving complex differential equations.