Solve the initial-value problem.
y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1
y = e^x (c1 + c2*x)
y' = e^x (c1 + c2 + c2*x)
Now plug in the initial conditions to get
e^2 (c1 + 2c2) = 0
e^2 (c1 + c2 + 2c2) = 1
Now just solve for c1 and c2
To solve the given initial-value problem, we can use a method called the method of undetermined coefficients or the characteristic equation.
1. Solve the characteristic equation:
The characteristic equation for the given second-order linear homogeneous differential equation is:
r^2 - 2r + 1 = 0
Factoring this equation, we get:
(r - 1)(r - 1) = 0
So, we have a repeated root r = 1.
2. Determine the general solution:
Since we have a repeated root, the general solution for the homogeneous equation is given by:
y(x) = (C1 + C2x)e^x
3. Apply the initial conditions to find specific values of the constants:
Using the initial condition y(2) = 0, we can substitute the values into the general solution and solve for the constants:
0 = (C1 + C2(2))e^2
0 = (C1 + 2C2)e^2
Using the initial condition y'(2) = 1, we differentiate the general solution and substitute the values into the derivative:
1 = (C2 + 2C2)e^2
1 = 3C2e^2
From the second equation, we can solve for C2:
C2 = e^(-2/3)
Substituting this value back into the first equation, we solve for C1:
0 = (C1 + 2e^(-2/3))e^2
C1 = -2e^(2/3)
4. Write the particular solution:
Now that we have the values for C1 and C2, we can write the particular solution that satisfies the initial conditions:
y(x) = (-2e^(2/3) + 2xe^(-2/3))e^x
Therefore, the solution to the initial-value problem is:
y(x) = (-2e^(2/3) + 2xe^(-2/3))e^x.