If `logy=tan^(-1)x,"then "(1+x^(2))y_(2)+(2x-1)y_(1)+4=`

Options

A. 0
B. `2logy`
C. 4
D. 1

Answer

To solve the problem, we start with the given equation:

\[ \log y = \tan^{-1} x \]

### Step 1: Differentiate both sides
We will differentiate both sides with respect to \( x \).

Using the chain rule on the left side, we have:

\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]

For the right side, the derivative of \( \tan^{-1} x \) is:

\[ \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2} \]

So, we can equate the derivatives:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{1 + x^2} \]

### Step 2: Rearranging the equation
Multiplying both sides by \( y(1 + x^2) \):

\[ (1 + x^2) \frac{dy}{dx} = y \]

### Step 3: Differentiate again
Now, we will differentiate both sides again. We apply the product rule on the left side:

\[ \frac{d}{dx}((1 + x^2) \frac{dy}{dx}) = (1 + x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} \]

The right side is simply \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} \]

So we have:

\[ (1 + x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = \frac{dy}{dx} \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ (1 + x^2) \frac{d^2y}{dx^2} + (2x - 1) \frac{dy}{dx} = 0 \]

### Step 5: Substitute into the original expression
Now, we need to find the expression:

\[ (1 + x^2) y_2 + (2x - 1) y_1 + 4 \]

Where \( y_2 = \frac{d^2y}{dx^2} \) and \( y_1 = \frac{dy}{dx} \). From our previous step, we have:

\[ (1 + x^2) y_2 + (2x - 1) y_1 = 0 \]

Thus, we can substitute this into our expression:

\[ (1 + x^2) y_2 + (2x - 1) y_1 + 4 = 0 + 4 = 4 \]

### Final Answer
Therefore, the final result is:

\[ (1 + x^2) y_2 + (2x - 1) y_1 + 4 = 4 \]