Question

Let y(x) be the solution to the initial value problem y ′′ - 10y ′ + 25y = 0, y(0) = 1, y′ (0) = 3. Find the value of y(1).

Answer 1

The solution to the initial value problem y '' - 10y' + 25y = 0 with y(0) = 1 and y'(0) = 3 is y(x) = (1 - 2x)e^(5x). The value of y at x = 1 is -e^5.

Step-by-step explanation:

To solve the initial value problem y '' - 10y' + 25y = 0, with initial conditions y(0) = 1 and y'(0) = 3, we first find the characteristic equation of the differential equation, which is r^2 - 10r + 25 = 0. This factors to (r - 5)^2 = 0, giving us a repeated root of r = 5. Therefore, the general solution is y(x) = (C1 + C2x)e^(5x), where C1 and C2 are constants.

Using the initial conditions, we can determine C1 and C2:

Substituting C1 and C2 back into the general solution, we get:

y(x) = (1 - 2x)e^(5x)

To find the value of y at x = 1, we plug in x = 1:

y(1) = (1 - 2(1))e^(5(1)) = -1e^5 = -e^5

The value of y(1) is -e^5.