Question

Y''+6y'+9y=0 y(0)=2, y'(0)=1

Answer 1

Характеристическое уравнение:

$\lambda^2+6\lambda+9=0=>(\lambda+3)^2=0=>\lambda_1=-3,d_{\lambda_1}=2$

$y=(C_1+C_2x)e^{-3x}\\ y(0)=2=>2=C_1\\ y'(0)=1=>-3C_1e^{0}+C_2e^{0}-3C_2*0e^{-3x}=1=>-3*2+C_2=1=>C_2=7\\ y_r=(2+7x)e^{-3x}$

Answer 2

The given differential equation is a second-order linear homogeneous ordinary differential equation. To solve it, we can assume a solution of the form y = e^(rt), where r is a constant to be determined.

Substituting this into the equation, we have:

y'' + 6y' + 9y = 0

(e^(rt))'' + 6(e^(rt))' + 9e^(rt) = 0

r^2e^(rt) + 6re^(rt) + 9e^(rt) = 0

Factoring out e^(rt), we get:

e^(rt)(r^2 + 6r + 9) = 0

For this equation to hold true, either e^(rt) = 0 (which is not possible) or (r^2 + 6r + 9) = 0.

The quadratic equation r^2 + 6r + 9 = 0 can be factored as (r + 3)(r + 3) = 0.

Therefore, we have a repeated root r = -3.

Since r = -3 is a repeated root, our general solution will involve multiplying by t as well. So the general solution is:

y(t) = (c1 + c2t)e^(-3t)

To find the particular solution that satisfies the initial conditions y(0) = 2 and y'(0) = 1, we can substitute these values into the general solution:

y(0) = (c1 + c2(0))e^(-3(0)) = c1 = 2

y'(0) = (c2 - 3c1)e^(-3(0)) = c2 - 3(2) = 1 c2 - 6 = 1 c2 = 7

Therefore, the particular solution that satisfies the initial conditions is:

y(t) = (2 + 7t)e^(-3t)

So the solution to the given differential equation with initial conditions y(0) = 2 and y'(0) = 1 is:

y(t) = (2 + 7t)e^(-3t)