Question

Y +2y'+y=0 y=7 y'=10 при x=0

Answer 1

Решение:
при у=7 у'=10
7+2*10+7=34
а при х=0 не знаю как решать уравнение,простите . ОТВЕТ ПРАВИЛЬНЫЙ.

Answer 2

To solve the given second-order linear differential equation, we can use the method of characteristic equations. Let's proceed step by step.

The differential equation is: y + 2y' + y = 0

Step 1: Find the characteristic equation. The characteristic equation is obtained by assuming the solution has the form y = e^(rx), where r is a constant.

Substituting this form into the differential equation, we get: e^(rx) + 2re^(rx) + e^(rx) = 0

Step 2: Simplify the characteristic equation. Combine like terms: 2re^(rx) + 2e^(rx) = 0

Factor out e^(rx): e^(rx)(2r + 2) = 0

Step 3: Solve the characteristic equation. Setting each factor equal to zero gives us: e^(rx) = 0 or 2r + 2 = 0

The equation e^(rx) = 0 does not have any real solutions, so we focus on the second equation: 2r + 2 = 0. Solving this equation for r, we find: 2r = -2 r = -1

Step 4: Find the general solution. Since we have a repeated root (r = -1), the general solution takes the form: y(x) = (C₁ + C₂x)e^(-x)

Step 5: Apply initial conditions. We are given the initial conditions y = 7 and y' = 10 when x = 0.

Substituting x = 0 into the general solution, we get: y(0) = (C₁ + C₂(0))e^(-0) = C₁ = 7

Differentiating y(x) with respect to x, we have: y'(x) = C₂e^(-x) - (C₁ + C₂x)e^(-x) y'(0) = C₂ - C₁ = 10

Using the value of C₁ obtained from the first initial condition, we have: C₂ - 7 = 10 C₂ = 10 + 7 C₂ = 17

Step 6: Final solution. With the values of C₁ and C₂, the particular solution to the initial value problem is: y(x) = (7 + 17x)e^(-x)