If the Particular Integral (P.I.) of the differential equation (D3 + 1...
Sin 2x), what is the general solution of the differential equation?
The general solution of a differential equation is the sum of the homogeneous solution and the particular integral.
The homogeneous solution of the differential equation (D^3 - 1)y = 0 can be found by assuming a solution of the form y = e^(mx), where m is a constant. Substituting this into the differential equation, we get (D^3 - 1)e^(mx) = 0. This simplifies to m^3e^(mx) - e^(mx) = 0. Factoring out e^(mx), we have e^(mx)(m^3 - 1) = 0. This equation holds if either e^(mx) = 0 (which is not possible) or m^3 - 1 = 0. Solving for m, we find m = 1, -1/2 + sqrt(3)/2, -1/2 - sqrt(3)/2.
Therefore, the homogeneous solution is y_h = C1e^x + C2e^((-1/2 + sqrt(3)/2)x) + C3e^((-1/2 - sqrt(3)/2)x), where C1, C2, and C3 are arbitrary constants.
The general solution of the differential equation is y = y_h + y_p. Substituting the given particular integral, we have y = C1e^x + C2e^((-1/2 + sqrt(3)/2)x) + C3e^((-1/2 - sqrt(3)/2)x) + 1/65 (cos 2x - sin 2x).