Let : (H) : x².y'' + 3x.y' - 4y = 0
Look for a solution of the form : y(x) = x^a
y' (x) = a. x^(a-1)
y'' (x) = a(a-1).x^(a-2)
y is a solution of (H) : a(a-1) + 3a - 4 = 0
Hence a = -1 - √5, or a = -1 + √5
A general solution for (H) is K₁ x^(-1-√5) + K₂ x^(-1+√5), where K₁ and K₂ are random constants.
Now let's find a specific solution y₀ for (E) : x².y'' + 3x.y' - 4y = e^x + 3
Let's try with y₀ = u(x).x^(-1-√5) + v(x).x^(-1+√5)
Express y'₀ and y''₀ in terms of x, and replace in (E). After some cancellation :
u''. x^(1-√5) + (1+√5) u'.x^(-√5) + v''. x^(1+√5) + (1+√5) v'.x^(√5) = e^x + 3
Let u' = - v'. x^(2√5) : u'' = - v''.x^(2√5) - 2√5 v'. x^(-1+2√5)
and you are left with :
v' = - (e^x + 3).x^(-√5) / 2√5 and u' = (e^x + 3).x^(√5) / 2√5
And then : y₀ = [ ∫u' ].x^(-1-√5) + [ ∫v' ].x^(-1+√5)
(2√5) y₀ = [ ∫(e^t + 3).t^(√5)dt ].x^(-1-√5) - [ ∫(e^t + 3).t^(-√5)dt ].x^(-1+√5)
(2√5) y₀ = [ ∫ e^t . t^(√5)dt ].x^(-1-√5) + 3/(1+√5) - [ ∫ e^t . t^(-√5)dt ].x^(-1+√5) - 3/(1-√5)
y₀ = [ ∫ e^t . t^(√5)dt ].x^(-1-√5)/(2√5) - [ ∫ e^t . t^(-√5)dt ].x^(-1+√5)/(2√5) - 3/4
And finally, y(x) = K₁ x^(-1-√5) + K₂ x^(-1+√5) + y₀ (x)