Question

Y"+3y'+2y=0 н.у y(0)=-1 y'(0)=3

Answer 1

Ответ:

y = C₁ e^(-2x) + C₂ e^(-x)  

y(0) = C₁ + C₂, C₁ + C₂ = - 1, C₁ = - 1 - C₂

y` = - 2*C₁e^(-2x) - C₂e^(-x) , y`(0) = 3

- 2*C₁ - C₂ = 3,  

- 2*(-1 - C₂) - C₂ = 3

2 + 2C₂ - C₂ = 3

C₂ = 1

C₁ = - 1 - 1 = - 2

y = e^(-2x) - 2e^(-x)  

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

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Answer 2

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients:

Y'' + 3Y' + 2Y = 0

To solve this differential equation, we can first find the characteristic equation, which is obtained by replacing Y'' with r^2, Y' with r, and Y with 1 in the equation:

r^2 + 3r + 2 = 0

Now, we solve this quadratic equation for r. We can factor it:

(r + 2)(r + 1) = 0

This equation has two solutions: r = -2 and r = -1.

Since the roots of the characteristic equation are distinct and real, the general solution for the homogeneous differential equation can be written as:

Y(t) = C1 * e^(-2t) + C2 * e^(-t)

Now, we can apply the initial conditions to find the values of the constants C1 and C2:

1. Initial condition: y(0) = -1 We have: -1 = C1 * e^(-2 * 0) + C2 * e^(-1 * 0) -1 = C1 + C2

2. Initial condition: y'(0) = 3 We need to find the derivative of Y(t) and then evaluate it at t = 0: Y'(t) = -2C1 * e^(-2t) - C2 * e^(-t) Y'(0) = -2C1 * e^(-2 * 0) - C2 * e^(-1 * 0) = -2C1 - C2

Now, we have a system of two equations:

1. -1 = C1 + C2
2. 3 = -2C1 - C2

You can solve this system of equations to find the values of C1 and C2. Here's how you can do it:

From the first equation, we can express C1 in terms of C2:

C1 = -1 - C2

Now, substitute this expression for C1 into the second equation:

3 = -2(-1 - C2) - C2

Simplify and solve for C2:

3 = 2 + 2C2 - C2 3 = 2 + C2

C2 = 3 - 2 C2 = 1

Now that you have the value of C2, you can find C1 using the first equation:

C1 = -1 - C2 C1 = -1 - 1 C1 = -2

So, the values of the constants are C1 = -2 and C2 = 1. Therefore, the particular solution to the initial value problem is:

Y(t) = -2 * e^(-2t) + e^(-t)