To solve the given second-order linear non-homogeneous differential equation:

$\frac{d^2 y}{dx^2} - 4y = e^{2x}$

We can solve this in two parts:

1. Solve the homogeneous equation: $\frac{d^2 y}{dx^2} - 4y = 0$
2. Find the particular solution for the non-homogeneous part: $\frac{d^2 y}{dx^2} - 4y = e^{2x}$

Step 1: Solve the homogeneous equation

The homogeneous equation is: $\frac{d^2 y}{dx^2} - 4y = 0$

This can be written as a characteristic equation: $r^2 - 4 = 0$

Solving for $r$: $r^2 = 4 \quad \Rightarrow \quad r = \pm 2$

So, the general solution to the homogeneous equation is: $y_h(x) = C_1 e^{2x} + C_2 e^{-2x}$

Step 2: Find the particular solution

To find the particular solution $y_p(x)$, we guess a solution of the form $y_p(x) = A x e^{2x}$ because the right-hand side is $e^{2x}$ and $e^{2x}$ already appears in the homogeneous solution.

Now, differentiate $y_p(x) = A x e^{2x}$:

$\frac{dy_p}{dx} = A e^{2x} + 2A x e^{2x}$

$\frac{d^2 y_p}{dx^2} = 2A e^{2x} + 2A e^{2x} + 4A x e^{2x} = 4A e^{2x} + 4A x e^{2x}$

Substitute $y_p(x)$ and its derivatives into the original equation:

$(4A e^{2x} + 4A x e^{2x}) - 4(A x e^{2x}) = e^{2x}$

Simplifying: $4A e^{2x} = e^{2x}$

Equating coefficients of $e^{2x}$: $4A = 1 \quad \Rightarrow \quad A = \frac{1}{4}$

Thus, the particular solution is: $y_p(x) = \frac{1}{4} x e^{2x}$

Step 3: General solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions: $y(x) = C_1 e^{2x} + C_2 e^{-2x} + \frac{1}{4} x e^{2x}$

Final Answer:

$y(x) = C_1 e^{2x} + C_2 e^{-2x} + \frac{1}{4} x e^{2x}$

Would you like more details on any specific step, or do you have any questions?

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Here are five related questions:

1. How do you find the characteristic equation for higher-order linear differential equations?
2. What are the conditions when the method of undetermined coefficients is applicable?
3. How would you solve this differential equation if the non-homogeneous term was a polynomial?
4. Can the method of variation of parameters be applied to this problem?
5. What happens if the right-hand side was $e^{3x}$ instead of $e^{2x}$?

Tip: Always verify the particular solution by substituting it back into the original equation.