Converting Cos and Sin to Exponential in Differential Equations

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SUMMARY

The discussion focuses on converting the trigonometric functions cosine and sine into their exponential forms for the differential equation y = Acos(kx) + Bsin(kx). The transformation utilizes Euler's identity, where cos(kx) is expressed as 1/2[e^(jkx) + e^(-jkx)] and sin(kx) as 1/(2j)[e^(jkx) - e^(-jkx)]. The key challenge is to correctly assign coefficients A and B in the exponential form, which requires establishing a system of linear equations to relate the coefficients from the original trigonometric equation to the new exponential equation.

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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 =Aejkx + 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[ ejx + e-jx ]
sin(x) = 1/(2j) [ ejx - e-jx ]

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

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