Find a general solution to the differential equation given below. Primes denote derivatives with respect to $t$.
\[
y^{\prime \prime}-3 y^{\prime}-4 y=0
\]
A general solution is $y(t)=$

Step 1 ：This is a second order homogeneous linear differential equation with constant coefficients. The general form of such an equation is \(ay'' + by' + cy = 0\). The general solution to such an equation is given by \(y(t) = C_1e^{m_1t} + C_2e^{m_2t}\), where \(m_1\) and \(m_2\) are the roots of the characteristic equation \(am^2 + bm + c = 0\) and \(C_1\) and \(C_2\) are arbitrary constants.

Step 2 ：In this case, the characteristic equation is \(m^2 - 3m - 4 = 0\). We can solve this equation to find the roots \(m_1\) and \(m_2\).

Step 3 ：The roots of the characteristic equation are \(m_1 = -1\) and \(m_2 = 4\). Therefore, the general solution to the differential equation is \(y(t) = C_1e^{-t} + C_2e^{4t}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step 4 ：Final Answer: \(\boxed{y(t) = C_1e^{-t} + C_2e^{4t}}\)