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Mathematics Question on Solution of Differential Equations

Solution of a differential equation (1+y2)dx=(tan⁡−1y−x)dy(1+ y2)dx=(\tan^{-1}y - x)dy is :

A

yetan⁡−1y=etan⁡−1y(tan⁡−1y+1)+Cye^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y + 1) + C

B

yetan⁡−1y=etan⁡−1y(tan⁡−1y−1)+Cye^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y - 1) + C

C

xetan⁡−1y=etan⁡−1y(tan⁡−1y+1)+Cxe^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y + 1) + C

D

xetan⁡−1y=etan⁡−1y(tan⁡−1y−1)+Cxe^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y - 1) + C

Answer

xetan⁡−1y=etan⁡−1y(tan⁡−1y+1)+Cxe^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y + 1) + C

Explanation

Solution

The correct option is (C):xetan⁡−1y=etan⁡−1y(tan⁡−1y−1)+Cxe^{\tan^{-1}y }= e^{\tan^{-1}y}(\tan^{-1}y - 1) + C