You seem to be confused about the solutions to ODEs.
For the equation y" - y = 0, with y(0) = 5 and y'(0) = 3, we have a homogeneous ODE for which we find the complementary solution by assuming y = e^(rx). Differentiating and substituting into the original ODE gives:
r^2 * e^(rx) - e^rx = 0
Dividing both sides by e^rx (which is always > 0, for any x), we obtain the characteristic polynomial
r^2 - 1 = 0 or r^2 = 1, which has solutions r = 1 and r = -1.
This means that e^x and e^-x are both solutions to the original ODE. Since the original ODE is linear, the sum of the individual solutions obtained from solving the characteristic polynomial must also be a solution, thus
y = C1*e^-x + C2 * e^x
By applying the initial conditions y(0) = 5 and y'(0) = 3, the coefficients of the general solution C1 and C2 can be determined.