Given that exp(2x) is a solution, we assume another solution of the form
y(x) = v(x) exp(2x) = v exp(2x)
with derivatives
y' = v' exp(2x) + 2v exp(2x)
y'' = v'' exp(2x) + 4v' exp(2x) + 4v exp(2x)
Substitute these into the equation:
(2x + 5) (v'' exp(2x) + 4v' exp(2x) + 4v exp(2x)) - 4 (x + 3) (v' exp(2x) + 2v exp(2x)) + 4v exp(2x) = 0
Each term contains a factor of exp(2x) that can be divided out:
(2x + 5) (v'' + 4v' + 4v) - 4 (x + 3) (v' + 2v) + 4v = 0
Expanding and simplifying eliminates the v term:
(2x + 5) v'' + (4x + 8) v' = 0
Substitute w(x) = v'(x) to reduce the order of the equation, and you're left with a linear ODE:
(2x + 5) w' + (4x + 8) w = 0
w' + (4x + 8)/(2x + 5) w = 0
I'll use the integrating factor method. The IF is
µ(x) = exp( ∫ (4x + 8)/(2x + 5) dx ) = exp(2x - log|2x + 5|) = exp(2x)/(2x + 5)
Multiply through the ODE in w by µ :
µw' + µ (4x + 8)/(2x + 5) w = 0
The left side is the derivative of a product:
[µw]' = 0
Integrate both sides:
∫ [µw]' dx = ∫ 0 dx
µw = C
Replace w with v', then integrate to solve for v :
exp(2x)/(2x + 5) v' = C
v' = C (2x + 5) exp(-2x)
∫ v' dx = ∫ C (2x + 5) exp(-2x) dx
v = C₁ (x + 3) exp(-2x) + C₂
Replace v with y exp(-2x) and solve for y :
y exp(-2x) = C₁ (x + 3) exp(-2x) + C₂
y = C₁ (x + 3) + C₂ exp(2x)
It follows that the second fundamental solution is y = x + 3. (The exp(2x) here is already accounted for as the first solution.)