Question

Y''+3y'+2y=0,x=0,y=1,y'=4

Answer 1

Ответ:

решение на фотографии

Answer 2

To solve the given second-order ordinary differential equation with the given initial conditions, we need to find the solution y(x). The equation is:

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

First, let's find the characteristic equation by assuming a solution of the form:

Y(x) = e^(rx)

where r is a constant to be determined. Substituting this into the differential equation, we get:

r^2 e^(rx) + 3r e^(rx) + 2 e^(rx) = 0

Divide the equation by e^(rx) to simplify:

r^2 + 3r + 2 = 0

Now, we can solve the quadratic equation for r:

r = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = 3, and c = 2:

r = (-(3) ± √(3^2 - 4 * 1 * 2)) / (2 * 1)

r = (-3 ± √(9 - 8)) / 2

r = (-3 ± √1) / 2

Now, we have two possible values for r:

1. r1 = (-3 + 1) / 2 = -1
2. r2 = (-3 - 1) / 2 = -2

The general solution for the differential equation is a linear combination of the solutions for r1 and r2:

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

Now, we apply the initial conditions to find the values of C1 and C2.

Given: x = 0, y = 1, y' = 4

Substitute x = 0 and y = 1 into the general solution:

1 = C1 * e^(0) + C2 * e^(0) 1 = C1 + C2

Next, differentiate the general solution with respect to x to find y':

y' = -C1 * e^(-x) - 2C2 * e^(-2x)

Now, substitute x = 0 and y' = 4 into the y' equation:

4 = -C1 * e^(0) - 2C2 * e^(0) 4 = -C1 - 2C2

Now, we have a system of two equations with two unknowns:

1. C1 + C2 = 1
2. -C1 - 2C2 = 4

We can solve this system of equations to find the values of C1 and C2.

Adding both equations:

C1 + C2 - C1 - 2C2 = 1 + 4

Simplifying:

-C2 = 5

Divide by -1:

C2 = -5

Now, substitute the value of C2 back into the first equation to find C1:

C1 + (-5) = 1

C1 = 1 + 5

C1 = 6

So, the values of C1 and C2 are:

C1 = 6 C2 = -5

The final solution for the given differential equation with the initial conditions is:

Y(x) = 6 * e^(-x) - 5 * e^(-2x)