Solve the IVP:
4y''(x)-4y'(x)-3y(x)=0
y'(0)=5
y(0)=1
So far I have y = (-1/2) and y=(3/2)
y(x) = (c1e^(3/2)(x))+(c^2e^(-1/2)(x))
c1= -1
c2 = 3
$4y^{\prime \prime} - 4 y^\prime - 3y = 0$
characteristic equation is
$4s^2 - 4s -3 = 0$
with roots $s=-\dfrac 1 2,~\dfrac 3 2$
and thus solution
$y(x) = c_1 e^{-x/2}+c_2 e^{3x/2}$
$y^\prime(x) = -\dfrac {c_1}{2} e^{-x/2} + \dfrac {3c_2}{2} e^{3x/2}$
$y^\prime(0) = -\dfrac{c_1}{2} + \dfrac{3c_2}{2} = 5$
$y(0) = c_1 + c_2 = 1$
solving we get
$c_1 = -\dfrac 7 4, ~ c_2 = \dfrac {11}{4}$
$y(x) = -\dfrac 7 4 e^{-x/2} + \dfrac {11}{4} e^{3x/2}$