Statement I y=asin x+bcos x is general solution of y''+y=0.
Statement II y=a sin x+b cos x is a trigonometic function.

Options

A. Statement I is true ,and Statement II is the correct explanation for Statement I.
B. Statement I is true, Statement II is true and Statement II is not the correct explanation for Statment I
C. Statement I is true, Statement II is false.
D. Statement I is false, Statement II is true.

Answer

To solve the problem, we need to verify the two statements given:

**Statement I:** \( y = a \sin x + b \cos x \) is the general solution of \( y'' + y = 0 \).

**Statement II:** \( y = a \sin x + b \cos x \) is a trigonometric function.

### Step-by-Step Solution:

1. **Define the function**:
Let \( y = a \sin x + b \cos x \).

2. **Find the first derivative**:
Differentiate \( y \) with respect to \( x \):
\[ y' = \frac{dy}{dx} = a \cos x - b \sin x \]

3. **Find the second derivative**:
Differentiate \( y' \) with respect to \( x \):
\[ y'' = \frac{d^2y}{dx^2} = -a \sin x - b \cos x \]

4. **Substitute into the differential equation**:
Now substitute \( y \) and \( y'' \) into the equation \( y'' + y = 0 \):
\[ y'' + y = (-a \sin x - b \cos x) + (a \sin x + b \cos x) \]
Simplifying this gives:
\[ y'' + y = 0 \]

5. **Conclusion for Statement I**:
Since \( y'' + y = 0 \) holds true, Statement I is confirmed to be true.

6. **Evaluate Statement II**:
The function \( y = a \sin x + b \cos x \) is clearly composed of sine and cosine functions, which are trigonometric functions. Thus, Statement II is also true.

7. **Relationship between Statements**:
While both statements are true, Statement II does not serve as a correct explanation for Statement I. Statement I is proven through the differential equation, while Statement II merely describes the nature of the function.

### Final Conclusion:
- Statement I is true.
- Statement II is true.
- However, Statement II does not correctly explain Statement I.