Find the general solution of the linear homogeneous differential equation $y^{\prime \prime}-4 y^{\prime}+5 y=0$.

Step 1 ：This is a second order homogeneous linear differential equation with constant coefficients. The general solution of such an equation is given by \(y = e^{rt}\), where \(r\) is a root of the characteristic equation \(r^2 - 4r + 5 = 0\). We need to find the roots of this equation.

Step 2 ：The roots of the characteristic equation are complex numbers \(2 - i\) and \(2 + i\). This means that the general solution of the differential equation is of the form \(y = e^{rt}\), where \(r\) is either of the roots.

Step 3 ：Since the roots are complex, the solution will involve sine and cosine functions. The general solution is given by \(y = e^{2t}(A\cos(t) + B\sin(t))\), where \(A\) and \(B\) are arbitrary constants.

Step 4 ：\(\boxed{\text{The general solution of the linear homogeneous differential equation } y^{\prime \prime}-4 y^{\prime}+5 y=0 \text{ is } y = e^{2t}(A\cos(t) + B\sin(t))}\)