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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

Corbin
4.6
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
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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