Find the general solution of the equation y''− 4y'+ 13y = 0.
Let us find the general solution of the differential equation "y''\u2212 4y'+ 13y = 0". The characteristic equation "k^2-4k+13=0" is equivalent to "(k-2)^2=-9," and hence has the roots "k_1=2+3i" and "k_2=2-3i." Therefore, the general solutions is of the form:
"y=e^{2x}(C_1\\cos (3x)+C_2\\sin(3x))."
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