Consider the DE which is linear with constant coefficients. d²y dx² First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is = 0 which has root Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: 3/2 dy 8 + 16y = x dx reduction of order. Y2 = ue4x Then (using the prime notation for the derivatives) = y₁ = So, plugging y2 into the left side of the differential equation, and reducing, we get y8y2 + 16y2 = U= So now our equation is exu" =x. To solve for u we need only integrate xe twice, using a as our first constant of integration and b as the second we get -4x Therefore y2 = general solution. to do the We knew from the beginning that e4 was a solution. We have worked out is that xe4 is another solution to the homogeneous equation, which is generally the case 2+4x when we have multiple roots. Then 24 is the particular 64 solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.