Find the complete general solution and a particular solution that satisfies the initial conditions:
2y'' + y' - 4y = 0
y(0)=0, y'(0)=1
The solution will be
c1 e^k1x + c2 e^k2x
where k1 and k2 are the roots of 2x^2+x-4=0.
x = (-1±√33)/4
So, y = c1 e(-1+√33)/4 x + c2 e(-1-√33)/4 x
y(0) = 0, so c1+c2=0
y'(0) = 1 so (-1+√33) c1 + (-1-√33) c2 = 1
c1 = 1/(2√33)
c2 = -1/(2√33)