Final answer:
The general solution to the differential equation y'' - 2y' + y = 0 is y(x) = c1*e^x + c2*xe^x. Given the initial conditions y(0) = 5 and y'(0) = 10, we find c1 = 5 and c2 = 5, leading to the specific solution y(x) = 5e^x + 5xe^x.
Step-by-step explanation:
The differential equation given is y'' - 2y' + y = 0. To find the fundamental set of solutions and the general solution, we need to solve this second-order linear homogeneous differential equation.
We start by finding the characteristic equation which is formed by replacing y'' with r^2, y' with r, and y with 1. This yields r^2 - 2r + 1 = 0, which factors to (r - 1)^2 = 0. This indicates a repeated root at r = 1.
The fundamental set of solutions for a repeated root is e^(rx) and xe^(rx). Thus, our fundamental set of solutions is {e^x, xe^x}, and the general solution is y(x) = c1*e^x + c2*xe^x with constants c1 and c2.
To solve the initial value problem with conditions y(0) = 5 and y'(0) = 10, we plug in x = 0 into our general solution and its derivative to find the values of c1 and c2.
- Using y(0) = 5, we get 5 = c1 + 0, so c1 = 5.
- For the derivative, y'(x) = c1*e^x + c2*e^x + c2*xe^x, and using y'(0) = 10, we get 10 = 5*e^0 + c2*e^0 hence c2 = 5.
The specific solution to the initial value problem is thus y(x) = 5e^x + 5xe^x.