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Mathematics Question on Continuity and differentiability

If log⁡y=mtan⁡−1x,\log y= m \tan^{-1} x, then

A

(1+x2)y2+(2x+m)y1=0(1 + x^2)y^2 + (2x + m)y_1 = 0

B

(1+x2)y2+(2x−m)y1=0(1 + x^2)y^2 + (2x - m)y_1 = 0

C

(1+x2)y2−(2x+m)y1=0(1 + x^2)y^2 - (2x + m)y_1 = 0

D

(1+x2)y2−(2x−m)y1=0(1 + x^2)y^2 - (2x - m)y_1 = 0

Answer

(1+x2)y2+(2x−m)y1=0(1 + x^2)y^2 + (2x - m)y_1 = 0

Explanation

Solution

We have, log⁡y=mtan⁡−1x\log y= m \tan^{-1} x
Differentating w.r.t. x, we get
1ydydx=m1+x2\frac{1}{y} \frac{dy}{dx} = \frac{m}{1+x^{2}}
or , y1(1+x2)=myy_{1} \left(1+x^{2}\right) =my
Again differentiating w.r.t, x, we get
y2(1+x2)+y1(2x)=my1y_{2}\left(1+x^{2}\right)+y_{1}\left(2x\right)=my_{1}
⇒(1+x2)y2+(2x−m)y1=0\Rightarrow \left(1+x^{2}\right) y_{2} +\left(2x -m\right)y_{1} = 0