We are tasked with solving the differential equation:

\[
(D^2 + 2D + 3)y = e^x
\]

Where \( D \) is the differential operator with respect to \( x \), so \( D^n y = \frac{d^n y}{dx^n} \).

### Step 1: Find the homogeneous solution

First, we solve the homogeneous equation:

\[
(D^2 + 2D + 3)y = 0
\]

This is a second-order linear differential equation. To solve it, we first find the characteristic equation by replacing \( D \) with \( r \) (as if the solution were \( e^{rx} \)):

\[
r^2 + 2r + 3 = 0
\]

Solving this quadratic equation:

\[
r = \frac{-2 \pm \sqrt{2^2 - 4(1)(3)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 12}}{2} = \frac{-2 \pm \sqrt{-8}}{2} = \frac{-2 \pm 2i\sqrt{2}}{2}
\]

\[
r = -1 \pm i\sqrt{2}
\]

Thus, the roots are complex: \( r = -1 + i\sqrt{2} \) and \( r = -1 - i\sqrt{2} \).

The general solution to the homogeneous equation is:

\[
y_h(x) = e^{-x}(C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x))
\]

### Step 2: Find the particular solution

Now, we solve the non-homogeneous equation:

\[
(D^2 + 2D + 3)y = e^x
\]

Since the right-hand side is \( e^x \), we try a particular solution of the form \( y_p = A e^x \), where \( A \) is a constant to be determined.

Apply the differential operator \( (D^2 + 2D + 3) \) to \( y_p = A e^x \):

\[
D y_p = A e^x, \quad D^2 y_p = A e^x
\]

So,

\[
(D^2 + 2D + 3)y_p = A e^x + 2A e^x + 3A e^x = (A + 2A + 3A)e^x = 6A e^x
\]

We want this to equal \( e^x \), so:

\[
6A e^x = e^x
\]

Thus, \( 6A = 1 \), and \( A = \frac{1}{6} \).

The particular solution is:

\[
y_p(x) = \frac{1}{6} e^x
\]

### Step 3: General solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(x) = y_h(x) + y_p(x) = e^{-x}(C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x)) + \frac{1}{6} e^x
\]

This is the general solution to the differential equation.

Would you like more details or have any questions about the solution? Here are 5 related questions for further practice:

1. Solve the differential equation \( (D^2 + 5D + 6) y = 0 \).
2. Find the particular solution of \( (D^2 + 2D + 5) y = e^{2x} \).
3. What is the solution to \( (D^2 - 4D + 4) y = x^2 e^{2x} \)?
4. Solve the homogeneous equation \( (D^2 - D + 2) y = 0 \).
5. What are the roots of the characteristic equation \( r^2 + 4r + 13 = 0 \)?

**Tip:** Always test your particular solution by plugging it back into the original equation to ensure it satisfies the non-homogeneous equation.

This page covers the detailed solution of the differential equation (D^2 + 2D + 3) y = e^x. Learn about the characteristic equation, finding both the homogeneous and particular solutions, and apply the method of undetermined coefficients. Suitable for advanced high school or college students.

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