Question

If $y = e^{\tan^{-1} x}$, then

• A. (1+x2) y2 =(2x- 1) y1=0
• B. (1+x2)y2 + 2xy=0
• C. (1-x2) y2 -y1=0
• D. (1+x2) y2 +3xy1+4y=0

Answer 1

The Correct Option is A

Given: \( y = e^{\tan^{-1} x} \)

Step 1: Find the first derivative \( y' \) using chain rule.

\[ y' = \frac{d}{dx} \left( e^{\tan^{-1} x} \right) = e^{\tan^{-1} x} \cdot \frac{d}{dx}(\tan^{-1} x) \]

\[ \Rightarrow y' = e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \]

But \( y = e^{\tan^{-1} x} \), so: \[ y' = \frac{y}{1 + x^2} \]

Step 2: Differentiate again to get the second derivative \( y'' \)

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

Use quotient rule: \[ y'' = \frac{(1 + x^2) \cdot y' - y \cdot 2x}{(1 + x^2)^2} \]

Now substitute \( y' = \frac{y}{1 + x^2} \):

\[ y'' = \frac{(1 + x^2) \cdot \frac{y}{1 + x^2} - 2xy}{(1 + x^2)^2} = \frac{y - 2xy}{(1 + x^2)^2} \]

\[ \Rightarrow y'' = \frac{y(1 - 2x)}{(1 + x^2)^2} \]

Step 3: Multiply both sides by \( (1 + x^2)^2 \) to eliminate the denominator: \[ (1 + x^2)^2 y'' = y(1 - 2x) \]

Step 4: Rearranging: \[ (1 + x^2)^2 y'' - y(1 - 2x) = 0 \]

Or bring all terms to LHS: \[ (1 + x^2)^2 y'' + y(2x - 1) = 0 \]

But this is not matching the given options yet. Let's try expressing the final form in terms of \( y'', y', y \):

Recall: \[ y' = \frac{y}{1 + x^2} \Rightarrow y(2x - 1) = (2x - 1)(1 + x^2)y' \]

So: \[ (1 + x^2)^2 y'' + (2x - 1)(1 + x^2)y' = 0 \]

Divide the whole equation by \( 1 + x^2 \): \[ (1 + x^2) y'' + (2x - 1) y' = 0 \]

Final Result: \[ (1 + x^2) y'' + (2x - 1) y' = 0 \]

This matches option (1):
\((1 + x^2) y_2 + (2x - 1) y_1 = 0\)

Answer:

\[ \boxed{(1 + x^2) y'' + (2x - 1) y' = 0} \]