To find the total solution for each of the given differential equations, we will solve the homogeneous equation first, and then find the particular solution using the method of undetermined coefficients.
1. y'' + 16y = 32t, y(0) = 3, y'(0) = -2
The homogeneous equation is y'' + 16y = 0. The characteristic equation is r^2 + 16 = 0, which gives us imaginary roots r = ±4i. Therefore, the homogeneous solution is y_h(t) = c1*cos(4t) + c2*sin(4t).
To find the particular solution, we assume a solution of the form y_p(t) = At + B. Plugging this into the original equation, we get:
2A + 16(At + B) = 32t
(16A)t + (16B + 2A) = 32t
Comparing coefficients, we get A = 2 and B = -1/2. Therefore, the particular solution is y_p(t) = 2t - 1/2.
The total solution is the sum of the homogeneous and particular solutions: y(t) = y_h(t) + y_p(t) = c1*cos(4t) + c2*sin(4t) + 2t - 1/2.
Using the initial conditions, we can solve for the constants c1 and c2:
y(0) = c1*cos(0) + c2*sin(0) + 2(0) - 1/2 = 3
c1 - 1/2 = 3
c1 = 7/2
y'(t) = -4c1*sin(4t) + 4c2*cos(4t) + 2
y'(0) = -4c1*sin(0) + 4c2*cos(0) + 2(0) = -2
-4c1 + 4c2 + 2 = -2
-14 + 4c2 = -2
4c2 = 12
c2 = 3
Therefore, the total solution to the differential equation is: y(t) = (7/2)*cos(4t) + 3*sin(4t) + 2t - 1/2.
2. y'' - 3y' + 2y = 4, y(0) = 1, y'(0) = 0
The homogeneous equation is y'' - 3y' + 2y = 0. The characteristic equation is r^2 - 3r + 2 = 0, which factors as (r - 1)(r - 2) = 0. Therefore, the homogeneous solution is y_h(t) = c1*e^t + c2*e^(2t).
To find the particular solution, we assume a solution of the form y_p(t) = At + B. Plugging this into the original equation, we get:
2A - 3(At + B) + 2(At + B) = 4
(2A - 3B)t + (2B - 3A) = 4t
Comparing coefficients, we get A = 1 and B = -2. Therefore, the particular solution is y_p(t) = t - 2.
The total solution is the sum of the homogeneous and particular solutions: y(t) = y_h(t) + y_p(t) = c1*e^t + c2*e^(2t) + t - 2.
Using the initial conditions, we can solve for the constants c1 and c2:
y(0) = c1*e^0 + c2*e^(2*0) + 0 - 2 = 1
c1 + c2 - 2 = 1
c1 + c2 = 3
y'(t) = c1*e^t + 2c2*e^(2t) + 1
y'(0) = c1*e^0 + 2c2*e^(2*0) + 1 = 0
c1 + 2c2 + 1 = 0
c1 + 2c2 = -1
Solving the system of equations, we get c1 = 1 and c2 = 2.
Therefore, the total solution to the differential equation is: y(t) = e^t + 2e^(2t) + t - 2.
3. y'' + 2y' + y = 3te^(-t), y(0) = 4, y'(0) = 2
The homogeneous equation is y'' + 2y' + y = 0. The characteristic equation is r^2 + 2r + 1 = 0, which gives us a repeated root r = -1. Therefore, the homogeneous solution is y_h(t) = c1*e^(-t) + c2*t*e^(-t).
To find the particular solution, we assume a solution of the form y_p(t) = (At + B)e^(-t). Plugging this into the original equation, we get:
2Ae^(-t) + 2(-Ate^(-t) - Be^(-t)) + (At + B)e^(-t) = 3te^(-t)
(2A + At) e^(-t) = 3te^(-t)
Comparing coefficients, we get A = 3/2 and B = -3/2. Therefore, the particular solution is y_p(t) = (3/2)t - (3/2).
The total solution is the sum of the homogeneous and particular solutions: y(t) = y_h(t) + y_p(t) = c1*e^(-t) + c2*t*e^(-t) + (3/2)t - (3/2).
Using the initial conditions, we can solve for the constants c1 and c2:
y(0) = c1*e^(-0) + c2*0*e^(-0) + (3/2)*0 - (3/2) = 4
c1 - (3/2) = 4
c1 = 11/2
y'(t) = -c1*e^(-t) + c2*e^(-t) - c2*t*e^(-t) + (3/2)
y'(0) = -c1*e^(-0) + c2*e^(-0) - c2*0*e^(-0) + (3/2) = 2
-c1 + c2 + (3/2) = 2
-c1 + c2 = (1/2)
Solving the system of equations, we get c1 = 11/2 and c2 = 7/2.
Therefore, the total solution to the differential equation is: y(t) = (11/2)*e^(-t) + (7/2)*t*e^(-t) + (3/2)t - (3/2).
4. y'' + 4y' + 4y = 6e^(-2t), y(0) = -2, y'(0) = 8
The homogeneous equation is y'' + 4y' + 4y = 0. The characteristic equation is r^2 + 4r + 4 = 0, which gives us a repeated root r = -2. Therefore, the homogeneous solution is y_h(t) = c1*e^(-2t) + c2*t*e^(-2t).
To find the particular solution, we assume a solution of the form y_p(t) = Ae^(-2t). Plugging this into the original equation, we get:
-4Ae^(-2t) + 4Ae^(-2t) + 4Ae^(-2t) = 6e^(-2t)
4Ae^(-2t) = 6e^(-2t)
Comparing coefficients, we get A = 3/2. Therefore, the particular solution is y_p(t) = (3/2)e^(-2t).
The total solution is the sum of the homogeneous and particular solutions: y(t) = y_h(t) + y_p(t) = c1*e^(-2t) + c2*t*e^(-2t) + (3/2)e^(-2t).
Using the initial conditions, we can solve for the constants c1 and c2:
y(0) = c1*e^(-2*0) + c2*0*e^(-2*0) + (3/2)*e^(-2*0) = -2
c1 + (3/2) = -2
c1 = -7/2
y'(t) = -2c1*e^(-2t) - 2c2*t*e^(-2t) + (3/2)*(-2)e^(-2t)
y'(0) = -2c1*e^(-2*0) - 2c2*0*e^(-2*0) + (3/2)*(-2)e^(-2*0) = 8
-2c1 + (3/2)*(-2) = 8
-2c1 - 3 = 8
-2c1 = 11
c1 = -11/2
Solving the system of equations, we get c1 = -11/2 and c2 = 0.
Therefore, the total solution to the differential equation is: y(t) = (-11/2)*e^(-2t) + (3/2)e^(-2t).
PH