Find the complete general solution and a particular solution that satisfies the initial conditions:

Question

2y'' + y' - 4y = 0
y(0)=0, y'(0)=1

Answer 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)