This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Write the characteristic equation. The given differential equation is a second-order linear homogeneous equation with constant coefficients: y'' + 2y' + y = 0. We assume a solution of the form y = e^rx. Then y' = re^rx and y'' = r^2 e^rx. Substituting these into the differential equation gives: r^2 e^rx + 2re^rx + e^rx = 0 Divide by e^rx (since e^rx ≠ 0): r^2 + 2r + 1 = 0 This is the characteristic equation. Step 2: Solve the characteristic equation. The characteristic equation is a quadratic equation. It can be factored as a perfect square: (r+1)^2 = 0 This yields a repeated real root: r = -1 Step 3: Write the general solution. For a second-order linear homogeneous differential equation with a repeated real root r, the general solution is of the form: y(x) = C_1 e^rx + C_2 x e^rx Substitute the root r = -1: y(x) = C_1 e^-x + C_2 x e^-x Here, C_1 and C_2 are arbitrary constants. Step 4: Apply the initial condition. We are given the initial condition y(0) = 2. Substitute x=0 and y(0)=2 into the general solution: 2 = C_1 e^-(0) + C_2 (0) e^-(0) 2 = C_1 (1) + C_2 (0)(1) 2 = C_1 So, C_1 = 2. Step 5: Write the particular solution. Substitute the value of C_1 back into the general solution. Since no condition for y'(0) is provided, C_2 remains an arbitrary constant. y(x) = 2e^-x + C_2 x e^-x The solution to the differential equation with the given initial condition is: y(x) = 2e^-x + C_2 x e^-x where C_2 is an arbitrary constant. That's 2 down. 3 left today — send the next one.