$$ \frac { d ^ { 2 } y } { d x ^ { 2 } } - 4 \frac { d y } { d x } + 5 y = c $$

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1. **Identify the differential equation:**

We are given
   $$
   \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = c,
   $$
   where $$ c $$ is a constant.

2. **Solve the homogeneous part:**

The homogeneous equation is
   $$
   \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = 0.
   $$
   
   - Form the characteristic equation:
     $$
     r^2 - 4r + 5 = 0.
     $$
     
   - Solve for $$ r $$ using the quadratic formula:
     $$
     r = \frac{4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm \sqrt{-4}}{2}.
     $$
     Since $$\sqrt{-4} = 2i$$, the roots are
     $$
     r = 2 \pm i.
     $$
     
   - Therefore, the general solution to the homogeneous equation is given by
     $$
     y_h = e^{2x}\left(C_1 \cos x + C_2 \sin x\right),
     $$
     where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants.

3. **Find a particular solution:**

Since the nonhomogeneous term is the constant $$ c $$, we assume a particular solution of the form
   $$
   y_p = A,
   $$
   where $$ A $$ is a constant.
   
   - Substitute $$ y_p = A $$ into the differential equation. Note that
     $$
     \frac{dy_p}{dx} = 0 \quad \text{and} \quad \frac{d^2 y_p}{dx^2} = 0.
     $$
   - Substitute into the equation:
     $$
     0 - 4(0) + 5A = c \quad \Longrightarrow \quad 5A = c.
     $$
   - Solve for $$ A $$:
     $$
     A = \frac{c}{5}.
     $$
     
   Hence, the particular solution is
   $$
   y_p = \frac{c}{5}.
   $$

4. **Combine the solutions:**

The general solution to the nonhomogeneous differential equation is the sum of the homogeneous and particular solutions:
   $$
   y = y_h + y_p = e^{2x}\left(C_1 \cos x + C_2 \sin x\right) + \frac{c}{5}.
   $$
The general solution to the differential equation is: $$ y = e^{2x}(C_1 \cos x + C_2 \sin x) + \frac{c}{5} $$ where $$ C_1 $$ and $$ C_2 $$ are constants.

\( \frac { d ^ { 2 } y } { d x ^ { 2 } } - 4 \frac { d y } { d x } + 5 y = c \)

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