Final answer:
The general solution to the differential equation y"-4y'+5y=0 is y(t)=e^{2t}(C_1 cos(t)+C_2 sin(t)), where C_1 and C_2 are constants determined by initial conditions.
Step-by-step explanation:
To find a general solution of the second-order linear homogeneous differential equation y"-4y'+5y=0, we can use the characteristic equation method.
The characteristic equation is obtained by substituting y=e^{rt} into the differential equation, where r is a constant. This leads to the characteristic equation r^2-4r+5=0. Solving this quadratic equation for r gives us complex roots because the discriminant (b^2-4ac) is negative.
In our case, the roots are r=2+i and r=2-i. Consequently, the general solution is of the form y(t)=e^{2t}(C_1 cos(t)+C_2 sin(t)), where C_1 and C_2 are constants determined by initial conditions.