Question

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

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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 substitutingy = e^{rx} into the homogeneous equation:

r^2e^{rx} - 4re^{rx} + 4e^{rx} = 0

Factoring oute^{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 atr = 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 ofy_{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 ofc_1 and c_2 . Given: y(0) = 1 and y'(0) = 0 .

Substitutingx = 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

Substitutingx = 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 forc_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}

We first find the complementary function (CF) by solving the associated homogeneous equation:

The auxiliary equation is obtained by substituting

Factoring out

Simplifying the quadratic equation:

This implies a repeated root at

To find the particular integral (PI), we use the method of undetermined coefficients. Let's assume the particular solution is of the form:

Now, let's find the first and second derivatives of

Substituting these derivatives into the original differential equation and simplifying, we get:

Comparing coefficients, we have:

Equating the coefficients of like powers of x:

Solving the equations, we find:

Therefore, the particular solution (PI) is:

The general solution (GS) is the sum of the complementary function (CF) and the particular integral (PI):

Using the initial conditions, we can find the values of

Substituting

Substituting

Solving for

Therefore, the solution to the differential equation is:

Answer:

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