The particular solution that satisfies the initial conditions is:
y = (-83/20) e^(2x) + (123/10) x e^(2x) + (1/4) e^(2x) + (12/5) cos(3x) + sin(3x)
To solve the given initial value problem, which is a second-order ordinary differential equation (ODE) with constant coefficients, we'll follow these steps:
Step 1: Find the homogeneous solution by solving the associated homogeneous equation:
y'' - 4y' + 4y = 0
The characteristic equation is:
r^2 - 4r + 4 = 0
Solving this quadratic equation, we get:
(r - 2)^2 = 0
r - 2 = 0
r = 2 (double root)
Therefore, the homogeneous solution is:
y_h = C1 e^(2x) + C2 x e^(2x), where C1 and C2 are constants.
Step 2: Find a particular solution for the non-homogeneous equation:
We need to find a particular solution for the equation:
y'' - 4y' + 4y = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
We can assume a particular solution of the form:
y_p = A e^(2x) + B cos(3x) + C sin(3x)
Taking derivatives:
y_p' = 2A e^(2x) - 3B sin(3x) + 3C cos(3x)
y_p'' = 4A e^(2x) - 9B cos(3x) - 9C sin(3x)
Substituting these into the non-homogeneous equation, we get:
4A e^(2x) - 9B cos(3x) - 9C sin(3x) - 4(2A e^(2x) - 3B sin(3x) + 3C cos(3x)) + 4(A e^(2x) + B cos(3x) + C sin(3x)) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Simplifying the equation, we have:
4A e^(2x) - 8A e^(2x) + 4B cos(3x) + 9B cos(3x) - 4C sin(3x) - 9C sin(3x) + 4A e^(2x) + 4B cos(3x) + 4C sin(3x) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Grouping the terms, we get:
(-4A + 8A + 4A) e^(2x) + (4B - 9B + 4B) cos(3x) + (-4C - 9C + 4C) sin(3x) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Simplifying further:
8A e^(2x) - 5B cos(3x) - 5C sin(3x) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Equating the coefficients of the like terms on both sides, we have:
8A = 2, -5B = -12, -5C = -5
Solving these equations, we find:
A = 1/4, B = 12/5, C = 1
Therefore, a particular solution is:
y_p = (1/4) e^(2x) + (12/5) cos(3x) + sin(3x)
Step 3: Find the general solution:
The general solution is given by the sum of the homogeneous and particular solutions:
y = y_h + y_p
= C1 e^(2x) + C2 x e^(2x) + (1/4) e^(2x) + (12/5) cos(3x) + sin(3x)
Step 4: Apply initial conditions to find the values of constants:
Using the initial conditions y(0) = -2 and y'(0) = 4:
At x = 0:
-2 = C1 + (1/4) + (12/5)
-2 = C1 + (17/4) + (12/5)
C1 = -2 - (17/4) - (12/5)
C1 = -83/20
Differentiating y with respect to x:
y' = 2C1 e^(2x) + 2C2 x e^(2x) + C2 e^(2x) - (36/5) sin(3x) + 3 cos(3x)
To solve the given initial value problem, which is a second-order ordinary differential equation (ODE) with constant coefficients, we'll follow these steps:
Step 1: Find the homogeneous solution by solving the associated homogeneous equation:
y'' - 4y' + 4y = 0
The characteristic equation is:
r^2 - 4r + 4 = 0
Solving this quadratic equation, we get:
(r - 2)^2 = 0
r - 2 = 0
r = 2 (double root)
Therefore, the homogeneous solution is:
y_h = C1 e^(2x) + C2 x e^(2x), where C1 and C2 are constants.
Step 2: Find a particular solution for the non-homogeneous equation:
We need to find a particular solution for the equation:
y'' - 4y' + 4y = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
We can assume a particular solution of the form:
y_p = A e^(2x) + B cos(3x) + C sin(3x)
Taking derivatives:
y_p' = 2A e^(2x) - 3B sin(3x) + 3C cos(3x)
y_p'' = 4A e^(2x) - 9B cos(3x) - 9C sin(3x)
Substituting these into the non-homogeneous equation, we get:
4A e^(2x) - 9B cos(3x) - 9C sin(3x) - 4(2A e^(2x) - 3B sin(3x) + 3C cos(3x)) + 4(A e^(2x) + B cos(3x) + C sin(3x)) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Simplifying the equation, we have:
4A e^(2x) - 8A e^(2x) + 4B cos(3x) + 9B cos(3x) - 4C sin(3x) - 9C sin(3x) + 4A e^(2x) + 4B cos(3x) + 4C sin(3x) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Grouping the terms, we get:
(-4A + 8A + 4A) e^(2x) + (4B - 9B + 4B) cos(3x) + (-4C - 9C + 4C) sin(3x) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Simplifying further:
8A e^(2x) - 5B cos(3x) - 5C sin(3x) = 2 e^(2x) - 12 cos(3x) - 5 sin(3x)
Equating the coefficients of the like terms on both sides, we have:
8A = 2, -5B = -12, -5C = -5
Solving these equations, we find:
A = 1/4, B = 12/5, C = 1
Therefore, a particular solution is:
y_p = (1/4) e^(2x) + (12/5) cos(3x) + sin(3x)
Step 3: Find the general solution:
The general solution is given by the sum of the homogeneous and particular solutions:
y = y_h + y_p
= C1 e^(2x) + C2 x e^(2x) + (1/4) e^(2x) + (12/5) cos(3x) + sin(3x)
Step 4: Apply initial conditions to find the values of constants:
Using the initial conditions y(0) = -2 and y'(0) = 4:
At x = 0:
-2 = C1 + (1/4) + (12/5)
-2 = C1 + (17/4) + (12/5)
C1 = -2 - (17/4) - (12/5)
C1 = -83/20
Differentiating y with respect to x:
y' = 2C1 e^(2x) + 2C2 x e^(2x) + C2 e^(2x) - (36/5) sin(3x) + 3 cos(3x)
At x = 0:
4 = 2C1 + C2
4 = 2(-83/20) + C2
4 = -83/10 + C2
C2 = 4 + 83/10
C2 = 123/10
Therefore, the particular solution that satisfies the initial conditions is:
y = (-83/20) e^(2x) + (123/10) x e^(2x) + (1/4) e^(2x) + (12/5) cos(3x) + sin(3x)
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