Solve the second order differential equation d^2y/dx^2 - 4dy/dx + 5y = 15cos(x), given that when x = 0, y = 1 and when x = 0, dy/dx = 0

As we know, a general solution for a given differential equation is the complimentary solution + the particular solution/integral, this case is no different. To solve for the complimentary solution, form the auxillary equation by ignoring all non-y terms, in this case the 15cos(x), and setting the LHS to equal 0. Define the variable K to be the variable for our auxillary equation and the DE becomes K^2 -4K + 5 = 0. Solving the quadratic yields roots 2+i and 2-i. We know that complimentary solutions appear in the form Ae^lx + Be^mx where l and m are the roots of the aux. equation, however by application of De Moivre's Theorem by splitting the exponents, and grouping terms, you'll see that the comp. solution appears as e^2x(Acos(x) + Bsin(x)), where A and B are constants to be found.
For the particular integral, as the RHS in the form of cos(x), let the particular solution y = ecos(x) + fsin(x), where e and f are also constants to be found. Differentiating once yields us fcos(x) - esin(x), and differentiating once more yields -ecos(x) - fsin(x). By substituting these in to the left hand side, we can then compare coefficients and we find that e and f are 15/8 and -15/8 respectively. Adding the complimentary solution to the particular integral gives us e^2x(Acos(x) + Bsin(x)) +15/8(cos(x)-sin(x)). Considering our boundary conditions, we can set the RHS to 1 and substitute x = 0 on the left to get 1 = A + 15/8, getting A = 7/8. By differentiating once, applying the product rule and chain rule we get 2e^2x(Acos(x) + Bsin(x)) + e^2x(Bcos(x) - Asin(x)) - 15/8cos(x) -15/8sin(x). By setting the RHS to 0 and substituting x = 0 on the LHS. We obtain 0 = 2A + B - 15/8. Knowing that A = -7/8 and solving for B, we get B = 29/8. Thus, by grouping terms to tidy up, we obtain the final solution of y = e^2x/8(29sin(x) - 7cos(x)) + 1/8(cos(x) - sin(x)).