Solve the differential equation $y'' - 4y' + 4y = e^{2x}$ using the method of undetermined coefficients.

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∻Solution by Steps∻ ‖ step 1 ⋮ We start with the differential equation $y'' - 4y' + 4y = e^{2x}$. The complementary solution is found by solving the homogeneous equation $y'' - 4y' + 4y = 0$, which has a characteristic equation $(r - 2)^2 = 0$. Thus, the complementary solution is $y_c(x) = c_2 e^{2x} x + c_1 e^{2x}$. ‖ ‖ step 2 ⋮ Next, we find a particular solution using the method of undetermined coefficients. We assume a particular solution of the form $y_p(x) = A x^2 e^{2x}$, where $A$ is a constant to be determined. We then compute $y_p'$, $y_p''$, and substitute these into the original equation. ‖ ‖ step 3 ⋮ After substituting, we equate coefficients to solve for $A$. This gives us the particular solution $y_p(x) = \frac{1}{2} e^{2x} x^2$. ‖ ‖ step 4 ⋮ The general solution is then $y(x) = y_c(x) + y_p(x) = c_2 e^{2x} x + c_1 e^{2x} + \frac{1}{2} e^{2x} x^2$. ‖ ‖ step 5 ⋮ To find the particular solution at $x = 0$, we evaluate $y(0) = c_2(0) + c_1(1) + 0 = c_1$. Since all values of $y$ are solutions, we conclude that $y(0) = 0$ gives us $c_1 = 0$. ‖ ∻Answer∻ ⚹ The general solution is $y(x) = c_2 e^{2x} x + \frac{1}{2} e^{2x} x^2$ with $c_1 = 0$. ⚹ ∻Key Concept∻ ⚹ The method of undetermined coefficients is used to find particular solutions to non-homogeneous linear differential equations. ⚹ ∻Explanation∻ ⚹ The solution combines the complementary and particular solutions, and we find specific constants based on initial conditions or evaluations. ⚹◊What is the general solution for the complementary equation associated with the differential equation $y'' - 4y' + 4y = 0$?⍭ Generate me a similar question◊

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