$$ y ^ { \prime \prime } + y = x $$

Question

UpStudy AI · Accepted Answer

To solve the differential equation:

$$
y'' + y = x
$$

we'll find the **general solution**, which consists of two parts:

1. **Homogeneous Solution ($$ y_h $$)**: Solution to the associated homogeneous equation $$ y'' + y = 0 $$.
2. **Particular Solution ($$ y_p $$)**: A specific solution to the non-homogeneous equation $$ y'' + y = x $$.

### Step 1: Solve the Homogeneous Equation

The homogeneous equation is:

$$
y'' + y = 0
$$

**Characteristic Equation:**
$$
r^2 + 1 = 0 \quad \Rightarrow \quad r = \pm i
$$

**General Homogeneous Solution:**
$$
y_h = C_1 \cos(x) + C_2 \sin(x)
$$
where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants.

### Step 2: Find a Particular Solution

We need a specific solution to the equation:

$$
y'' + y = x
$$

Since the non-homogeneous term is a polynomial ($$ x $$), we'll try a particular solution of the form:

$$
y_p = A x + B
$$
where $$ A $$ and $$ B $$ are constants to be determined.

**Differentiate $$ y_p $$:**
$$
y_p' = A
$$
$$
y_p'' = 0
$$

**Substitute $$ y_p $$ into the differential equation:**
$$
0 + (A x + B) = x
$$

**Equate coefficients:**
$$
A x + B = x
$$

This gives us:
$$
A = 1 \quad \text{and} \quad B = 0
$$

So, the particular solution is:
$$
y_p = x
$$

### Step 3: Write the General Solution

Combine the homogeneous and particular solutions:

$$
y = y_h + y_p = C_1 \cos(x) + C_2 \sin(x) + x
$$

### Final Answer

$$
\boxed{ y(x) = C_1 \cos(x) + C_2 \sin(x) + x }
$$
where $$ C_1 $$ and $$ C_2 $$ are constants determined by initial conditions.
The general solution to the differential equation $$ y'' + y = x $$ is: $$ y(x) = C_1 \cos(x) + C_2 \sin(x) + x $$ where $$ C_1 $$ and $$ C_2 $$ are constants.

\( y ^ { \prime \prime } + y = x \)

Real Tutor Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions