Show that the general solution of the differential equation
4z'' - 4z' - 3z = cos2x takes the expression: z = c1e^(3x/2) + c2e^(-x/2) - (19/425)cos2x - (8/425)sin2x
Solution.
"4z''-4z'-3z=\\cos{2x}."This is a linear differential equation of the second-order with constant coefficients. The general solution of this equation is found as the sum of the general solution "\\tilde{z}" of the corresponding homogeneous equation and some particular integral solution "z_*\u200b" of the inhomogeneous equation:"zy=\\tilde{z}+z_*."
Сompose and solve the characteristic equation:
Therefore, "\\tilde{z}=C_1e^{\\lambda_1x}+C_2e^{\\lambda_2x}." So,
Since the right-hand side contains a "\\cos{2x}" , we find a particular integral solution "z_*" in the form:
Using the method of undetermined coefficients we find that "A=-\\frac{19}{425}, B=-\\frac{8}{425}."
So, the particular integral solution of equation "4z''-4z'-3z=\\cos{2x}" is
And the general solution of the same equation is
Answer. "z=C_1e^{-\\frac{1}{2}x}+C_2e^{\\frac{3}{2}x}-\\frac{19}{425}\\cos{2x}-\\frac{8}{425}\\sin{2x}."
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