$$ {\displaylines y^{\prime\prime}+y=x^2+3+\cos x的通解\\ \placeholder{}} $$

To solve the differential equation $ y^{\prime\prime} + y = x^2 + 3 + \cos x $, we need to find the general solution. This equation is non-homogeneous, so the general solution is composed of two parts: the **complementary solution** (solving the homogeneous equation) and the **particular solution** (accounting for the non-homogeneous part).

### Step 1: Solve the Homogeneous Equation

The homogeneous equation is:
\[
y^{\prime\prime} + y = 0
\]
The characteristic equation for this is:
\[
r^2 + 1 = 0
\]
Solving for $ r $:
\[
r = \pm i
\]
Thus, the complementary solution is:
\[
y_c = c_1 \cos x + c_2 \sin x
\]
where $ c_1 $ and $ c_2 $ are constants.

### Step 2: Find a Particular Solution

We now find the particular solution for the non-homogeneous equation:
\[
y^{\prime\prime} + y = x^2 + 3 + \cos x
\]
We break the right-hand side into two parts:
1. $ x^2 + 3 $
2. $ \cos x $

#### Particular Solution for $ x^2 + 3 $

Assume a particular solution of the form:
\[
y_p^{(1)} = Ax^2 + Bx + C
\]
Substituting this into the equation $ y^{\prime\prime} + y = x^2 + 3 $, we find the coefficients $ A $, $ B $, and $ C $.

The second derivative of $ y_p^{(1)} $ is:
\[
y_p^{(1)\prime\prime} = 2A
\]
Substitute $ y_p^{(1)} $ and $ y_p^{(1)\prime\prime} $ into the equation:
\[
2A + (Ax^2 + Bx + C) = x^2 + 3
\]
Equating coefficients of like powers of $ x $, we get:
\[
A = 1, \quad B = 0, \quad 2A + C = 3 \implies C = 1
\]
So, the particular solution for this part is:
\[
y_p^{(1)} = x^2 + 1
\]

#### Particular Solution for $ \cos x $

Since $ \cos x $ is already a solution to the homogeneous equation, we assume a particular solution of the form:
\[
y_p^{(2)} = x(A \cos x + B \sin x)
\]
Substitute this into $ y^{\prime\prime} + y = \cos x $, and solve for $ A $ and $ B $.

The second derivative of $ y_p^{(2)} $ is more involved, but after solving, we find:
\[
y_p^{(2)} = -\frac{1}{2}x \sin x
\]

### Step 3: General Solution

The general solution is the sum of the complementary and particular solutions:
\[
y = y_c + y_p = (c_1 \cos x + c_2 \sin x) + (x^2 + 1) - \frac{1}{2}x \sin x
\]

Thus, the general solution is:
\[
y = c_1 \cos x + \left( c_2 - \frac{1}{2}x \right) \sin x + x^2 + 1
\]

---

Would you like further details or have any questions?

Here are five related questions:
1. How do you find the complementary solution of a second-order differential equation?
2. What is the method of undetermined coefficients in solving differential equations?
3. Why is it necessary to modify the particular solution for $ \cos x $?
4. Can you explain how to handle the second derivative of a product like $ x \cos x $?
5. How would the solution change if the right-hand side were $ e^x $ instead of $ \cos x $?

**Tip:** Always check if your particular solution is already part of the complementary solution to avoid redundancy.

$$ {\displaylines y^{\prime\prime}+y=x^2+3+\cos x的通解\ \placeholder{}} $$

Explore the solution to the second-order differential equation y'' + y = x^2 + 3 + cos x, which involves solving the homogeneous and non-homogeneous parts using the method of undetermined coefficients. Detailed steps include finding the complementary and particular solutions. Suitable for advanced high school or college students.

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