Consider the DEdx 2 d 2 y −16 dxdy +64y=x
which is linear with constant coefficients. 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: to do reduction of order.
y 2 =ue ^8x
Then (using the prime notation for the derivatives)
y 2′ =
y 2′′ =
So, plugging y 2 into the left side of the differential equation, and reducing, we get y 2′′ −16y 2′ +64y 2 =So now our equation is e 8x u ′′ =x. To solve for u we need only integrate xe −8x twice, using a as our first constant of integration and b as the second we get
u=
Therefore y 2 = , the general solution. We knew from the beginning that e 8x was a solution. We have worked out is that xe 8x is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 512 2+8x is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.
