Which of the following functions are solutions of the differential equation y'' + y = 9 sin(x)? (Select all that apply.)

y = 9 sin(x)

y = − 
9
2
x cos(x)

y = 
9
2
x sin(x)

y = 9x sin(x) − 6x cos(x)
     

y = 9 cos(x)


To determine which of the given functions are solutions to the differential equation:

\[
y'' + y = 9 \sin(x)
\]

we will check each function by calculating their second derivatives and substituting into the equation.

### 1. \( y = 9 \sin(x) \)

- First derivative: \( y' = 9 \cos(x) \)
- Second derivative: \( y'' = -9 \sin(x) \)

Substitute into the equation:

\[
y'' + y = -9 \sin(x) + 9 \sin(x) = 0 \neq 9 \sin(x)
\]

Thus, **this is not a solution**.

---

### 2. \( y = -\frac{9}{2} x \cos(x) \)

- First derivative: 
  \[
  y' = -\frac{9}{2} (\cos(x) - x \sin(x)) = -\frac{9}{2} \cos(x) + \frac{9}{2} x \sin(x)
  \]
- Second derivative: 
  \[
  y'' = -\frac{9}{2} (-\sin(x) - \sin(x) - x \cos(x)) = 9 \sin(x) - \frac{9}{2} x \cos(x)
  \]

Substitute into the equation:

\[
y'' + y = 9 \sin(x) - \frac{9}{2} x \cos(x) + \left( -\frac{9}{2} x \cos(x) \right) = 9 \sin(x)
\]

Thus, **this is a solution**.

---

### 3. \( y = \frac{9}{2} x \sin(x) \)

- First derivative:
  \[
  y' = \frac{9}{2} (\sin(x) + x \cos(x)) = \frac{9}{2} \sin(x) + \frac{9}{2} x \cos(x)
  \]
- Second derivative:
  \[
  y'' = \frac{9}{2} (\cos(x) + \cos(x) - x \sin(x)) = 9 \cos(x) - \frac{9}{2} x \sin(x)
  \]

Substitute into the equation:

\[
y'' + y = 9 \cos(x) - \frac{9}{2} x \sin(x) + \frac{9}{2} x \sin(x) = 9 \cos(x)
\]

Thus, **this is not a solution**.

---

### 4. \( y = 9x \sin(x) - 6x \cos(x) \)

- First derivative:
  \[
  y' = 9 (\sin(x) + x \cos(x)) - 6 (\cos(x) - x \sin(x)) = 9 \sin(x) + 9x \cos(x) - 6 \cos(x) + 6x \sin(x)
  \]
- Second derivative:
  \[
  y'' = 9 (\cos(x) + \cos(x) - x \sin(x)) - 6 (-\sin(x) - \sin(x) - x \cos(x)) = 18 \cos(x) - 9x \sin(x) + 12 \sin(x) + 6x \cos(x)
  \]

Substitute into the equation:

\[
y'' + y = (18 \cos(x) - 9x \sin(x) + 12 \sin(x) + 6x \cos(x)) + (9x \sin(x) - 6x \cos(x)) = 18 \cos(x) + 12 \sin(x)
\]

Thus, **this is not a solution**.

---

### 5. \( y = 9 \cos(x) \)

- First derivative: \( y' = -9 \sin(x) \)
- Second derivative: \( y'' = -9 \cos(x) \)

Substitute into the equation:

\[
y'' + y = -9 \cos(x) + 9 \cos(x) = 0 \neq 9 \sin(x)
\]

Thus, **this is not a solution**.

---

### Conclusion

The only solution to the differential equation \( y'' + y = 9 \sin(x) \) is:

- \( y = -\frac{9}{2} x \cos(x) \)

Would you like more details or have any questions?

---

Here are 5 related questions to explore further:
1. How do you verify a solution to a differential equation?
2. What are general methods for solving second-order linear differential equations?
3. What role do trigonometric functions play in solving differential equations?
4. How does the principle of superposition apply to differential equations?
5. What are homogeneous and particular solutions in differential equations?

**Tip:** Always ensure the second derivative is carefully computed when solving differential equations to avoid mistakes.

Which of the following functions are solutions of the differential equation y'' + y = 9 sin(x)? (Select all that apply.) y = 9 sin(x), y = −9/2 x cos(x), y = 9/2 x sin(x), y = 9x sin(x) − 6x cos(x), y = 9 cos(x)

Discover how to solve the second-order differential equation y'' + y = 9 sin(x) by evaluating different function solutions. Explore detailed steps for checking solutions, including first and second derivatives, using trigonometric identities. Suitable for students studying calculus and differential equations.

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