Question

Consider the differential equation \( y'' + 2y' + y = 0 \). If \( y(0) = 0 \) and \( y'(0) = 1 \), then the value of \( y(2) \) is:

Hint:

For second-order linear differential equations with constant coefficients, solve the characteristic equation to find the general solution. Use initial conditions to determine the constants.

Answer 1

Step 1: Solve the differential equation.
We start with the second-order linear differential equation: \[ y'' + 2y' + y = 0 \] This is a standard form of a differential equation with constant coefficients. The characteristic equation is: \[ r^2 + 2r + 1 = 0 \] Solving for \( r \), we find: \[ r = -1 \] Thus, the general solution to the differential equation is: \[ y(t) = C_1 e^{-t} + C_2 t e^{-t} \] Step 2: Apply initial conditions.
Using the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \), we substitute into the general solution and its derivative: \[ y(0) = C_1 = 0 \] \[ y'(t) = -C_2 e^{-t} + C_2 t e^{-t} \] Substituting \( y'(0) = 1 \): \[ C_2 = 1 \] Step 3: Final solution.
Thus, the solution is: \[ y(t) = t e^{-t} \] Step 4: Find \( y(2) \).
Substituting \( t = 2 \): \[ y(2) = 2 e^{-2} \] Step 5: Conclusion.
Thus, the value of \( y(2) \) is approximately 0.2707.