Find the general solution of the differential equation d^2y/dx^2 - 2(dy/dx) = 26sin(3x)

2 Parts to this question - requires a complimentary function (treating equation as if right-hand side (RHS) = 0) and a particular integral (the general solution for where the RHS = 26sin(3x) )
1) The complimentary function - find the auxiliary equation. let y = constant, dy/dx = m, d^2y/dx^2 = m^2 This gives m^2 - 2m = 0 -> m(m-2) = 0 -> m = 0, m = 22) Choose the correct general equation given the roots. As there are 2 real roots the general equation is y = Ae^(alpha x) + Be^(Beta x) (Explained in video)3) Therefore substituting in m = 0 and m = 2 for alpha and beta gives -> y = Ae^(2x) + Be^(0x) -> y=Ae^(2x) + B
Now find the particular integral. Equation no longer homogeneous (i.e= 0)and need to account for the 26sin(3x) on the RHS of the equation
4) Let y = Lambda(sin3x) + Mu(cos3x) dy/dx = 3Lambda(cos3x) - 3Mu(sin3x) d^2y/dx^2 = -9Lamba(sin3x) - 9Mu(cos3x)
5) Substitute back into inital equation and solve for Lambda and Mu;
Initial equation: d^2y/dx^2 - 2(dy/dx) = 26sin(3x)Substituting in: (-9Lamba(sin3x) - 9Mu(cos3x)) - 2(3Lambda(cos3x) - 3Mu(sin3x)) = 26sin3x
sin3x : -9Lambda + 6Mu = 26cos3x: -9Mu - 6Lambda = 0Solving simultaneous equations gives : Lambda = - 2, Mu = 4/3
General solution = Complimentary Function + Particular Integral
Therefore y=Ae^(2x) + B - 2sin3x + 4/3cos3x
Recap;
2 parts to general solution - complimentary function and particular integral1) CF - Treat as RHS = 0. Find auxiliary equation and use appropriate roots equation2) PI - find using general forms (i.e with lambda and mu, not numbers)3) Combine CF and PI for general solution
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