Final answer:
To solve the initial value problem 2y′′−3y′−2y=0, with y(0)=1 and y′(0)=−β, we can find the characteristic equation by assuming a solution of the form y=e^(rx). The general solution to the differential equation is y=c1e^(2x)+c2e^(-x), where c1 and c2 are constants. Solving the initial conditions gives the solution y=(1+β)e^(2x)/3+(2-β)e^(-x)/3.
Step-by-step explanation:
To solve the initial value problem 2y′′−3y′−2y=0, with y(0)=1 and y′(0)=−β, we can find the characteristic equation by assuming a solution of the form y=e^(rx). Substituting this into the differential equation, we get r^2-3r-2=0. Solving this quadratic equation, we find r=2 and r=-1.
The general solution to the differential equation is y=c1e^(2x)+c2e^(-x), where c1 and c2 are constants. We can use the initial conditions y(0)=1 and y′(0)=−β to find the values of c1 and c2. Plugging in x=0 and y(0)=1, we get c1+c2=1. Plugging in x=0 and y′(0)=−β, we get 2c1-c2=-β.
Solving these two equations simultaneously gives c1=(1+β)/3 and c2=(2-β)/3. Therefore, the solution to the initial value problem is y=(1+β)e^(2x)/3+(2-β)e^(-x)/3.