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y’’ -4y’ +4y = (12x^2 -6x)e^2x Y(0)= 1 Y’(0)=0 Y(x)=c1y1+c2y2+yp

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Answer to a math question y’’ -4y’ +4y = (12x^2 -6x)e^2x Y(0)= 1 Y’(0)=0 Y(x)=c1y1+c2y2+yp

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Corbin
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To solve the given second-order linear homogeneous ordinary differential equation:
y'' - 4y' + 4y = (12x^2 - 6x)e^{2x}
We first find the complementary function (CF) by solving the associated homogeneous equation:
y'' - 4y' + 4y = 0
The auxiliary equation is obtained by substituting y = e^{rx} into the homogeneous equation:
r^2e^{rx} - 4re^{rx} + 4e^{rx} = 0
Factoring out e^{rx}:
e^{rx}(r^2 - 4r + 4) = 0
Simplifying the quadratic equation:
r^2 - 4r + 4 = (r-2)^2 = 0
This implies a repeated root at r = 2, so the complementary function (CF) is:
y_{CF} = (c_1 + c_2x)e^{2x}

To find the particular integral (PI), we use the method of undetermined coefficients. Let's assume the particular solution is of the form:
y_{PI} = Ax^2e^{2x} + Bxe^{2x}
Now, let's find the first and second derivatives of y_{PI}:
y'_{PI} = (2Ax^2e^{2x} + 2Axe^{2x}) + (Be^{2x} + 2Bxe^{2x})
y''_{PI} = (4Ax^2e^{2x} + 8Axe^{2x} + 2Ae^{2x}) + (2Be^{2x} + 4Be^{2x} + 2Bxe^{2x})
Substituting these derivatives into the original differential equation and simplifying, we get:
(4Ax^2 + 12Bx + 12A - 6B)e^{2x} = (12x^2 - 6x)e^{2x}

Comparing coefficients, we have:
4Ax^2 + 12Bx + 12A - 6B = 12x^2 - 6x
Equating the coefficients of like powers of x:
4A = 12 \quad \text{(coefficient of }x^2)
12B + 12A - 6B = -6 \quad \text{(coefficient of }x)

Solving the equations, we find:
A = 3
B = -1

Therefore, the particular solution (PI) is:
y_{PI} = 3x^2e^{2x} - xe^{2x}

The general solution (GS) is the sum of the complementary function (CF) and the particular integral (PI):
y_{GS} = y_{CF} + y_{PI} = (c_1 + c_2x)e^{2x} + 3x^2e^{2x} - xe^{2x}

Using the initial conditions, we can find the values of c_1 and c_2. Given: y(0) = 1 and y'(0) = 0.

Substituting x = 0 and y = 1 into the general solution (GS):
y_{GS}(0) = (c_1 + c_2 \cdot 0)e^{2 \cdot 0} + 3 \cdot 0^2 e^{2 \cdot 0} - 0 \cdot e^{2 \cdot 0} = c_1 = 1

Substituting x = 0 and y' = 0 into the general solution (GS):
y'_{GS}(0) = (c_2)e^{2 \cdot 0} + 0 - 1 \cdot e^{2 \cdot 0} = c_2 - 1 = 0

Solving for c_2, we get:
c_2 = 1

Therefore, the solution to the differential equation is:
y(x) = (1 + x)e^{2x} + 3x^2e^{2x} - xe^{2x}

Answer: y(x) = (1 + x + 3x^2 - x)e^{2x} = (1 + 2x + 3x^2)e^{2x}

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