Find the general solution of: y'' - 4y' + $4y = e^(2x)/x²$ using variation of parameters.

Step 1: Find the complementary solution

First, we find the complementary solution $y_c$ by solving the homogeneous equation $y'' - 4y' + 4y = 0$. We form the characteristic equation $m^2 - 4m + 4 = 0$, which factors as $(m-2)^2 = 0$. This gives us a repeated root $m=2$. Therefore, the complementary solution is $y_c = c_1e^{2x} + c_2xe^{2x}$.

Step 2: Identify $y_1$, $y_2$ and $f(x)$

From the complementary solution, we identify $y_1 = e^{2x}$ and $y_2 = xe^{2x}$. The non-homogeneous term $f(x)$ is $e^{2x}/x^2$.

Step 3: Calculate the Wronskian

Next, we calculate the Wronskian $W(y_1, y_2)$. The Wronskian is the determinant of the matrix formed by $y_1$, $y_2$ and their first derivatives. After computing the determinant, we find that the Wronskian is $e^{4x}$.

Step 4: Calculate $u_1'$ and $u_2'$

Using the formulas for variation of parameters, we calculate $u_1'$ and $u_2'$. We substitute $y_1$, $y_2$, $f(x)$, and $W$ into the formulas. After simplification, we get $u_1' = -1/x$ and $u_2' = 1/x^2$.

Step 5: Integrate to find $u_1$ and $u_2$

Now, we integrate $u_1'$ and $u_2'$ to find $u_1$ and $u_2$. The integral of $-1/x$ is $-\ln|x|$, and the integral of $1/x^2$ is $-1/x$. We omit the constants of integration for the particular solution.

Step 6: Form the particular solution $y_p$

The particular solution $y_p$ is given by $u_1 y_1 + u_2 y_2$. Substituting the expressions for $u_1$, $u_2$, $y_1$, and $y_2$, we get $y_p = -e^{2x}\ln|x| - e^{2x}$.

Step 7: Write the general solution

Finally, the general solution is the sum of the complementary solution and the particular solution, $y = y_c + y_p$. We combine the results to get the final general solution.

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