Verify that the function π¦ = π1π (βπ+2π)π₯ + π2π (βπβ2π)π₯ is a solution to π¦ β²β² + 2ππ¦ β² + (π 2 + 4)π¦ = 0.
"y'=(-k+2i)c_1e^{(-k+2i)x}+(-k-2i)c_2e^{(-k-2i)x}"
"y''=(-k+2i)^2c_1e^{(-k+2i)x}+(-k-2i)^2c_2e^{(-k-2i)x}"
"y''+ 2\ud835\udc58y'+ (k^2+ 4)\ud835\udc66"
"=(-k+2i)^2c_1e^{(-k+2i)x}+(-k-2i)^2c_2e^{(-k-2i)x}"
"+2k(-k+2i)c_1e^{(-k+2i)x}+2k(-k-2i)c_2e^{(-k-2i)x}"
"+ (k^2+ 4)c_1e^{(-k+2i)x}+ (k^2+ 4)c_2e^{(-k-2i)x}"
"=(k^2-4ik-4-2k^2+4ik+k^2+4)c_1e^{(-k+2i)x}"
"+(k^2-4ik-4-2k^2+4ik+k^2+4)c_2e^{(-k-2i)x}"
"=0, x\\in \\Complex"
Therefore the function
is a solution to
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