1. The given equation is probably supposed to read
y'' - 2y' - 3y = 64x exp(-x)
First consider the homogeneous equation,
y'' - 2y' - 3y = 0
which has characteristic equation
r² - 2r - 3 = (r - 3) (r + 1) = 0
with roots r = 3 and r = -1. Then the characteristic solution is
and we let y₁ = exp(3x) and y₂ = exp(-x), our fundamental solutions.
Now we use variation of parameters, which gives a particular solution of the form
where
and W(y₁, y₂) is the Wronskian determinant of the two fundamental solutions. This is
Then we find
so it follows that the particular solution is
and so the general solution is
2. I'll again assume there's typo in the equation, and that it should read
y''' - 6y'' + 11y' - 6y = 2x exp(-x)
Again, we consider the homogeneous equation,
y''' - 6y'' + 11y' - 6y = 0
and observe that the characteristic polynomial,
r³ - 6r² + 11r - 6
has coefficients that sum to 1 - 6 + 11 - 6 = 0, which immediately tells us that r = 1 is a root. Polynomial division and subsequent factoring yields
r³ - 6r² + 11r - 6 = (r - 1) (r² - 5r + 6) = (r - 1) (r - 2) (r - 3)
and from this we see the characteristic solution is
For the particular solution, I'll use undetermined coefficients. We look for a solution of the form
whose first three derivatives are
Substituting these into the equation gives
It follows that -24a = 2 and 26a - 24b = 0, so that a = -1/12 = -12/144 and b = -13/144, so the particular solution is
and the general solution is