Question

The differential equation

(1 + y²) + $\left( x - e ^ { \tan ^ { - 1 } y } \right)$ $\frac{dy}{dx}$= 0, has its solution

• A. (1 + y²) + $\left( x - e ^ { \tan ^ { - 1 } y } \right)$ $\frac{dy}{dx}$= 0, has its solution
• B. (x –2) = $Ke^{\tan^{- 1}y}$
• C. $2xe^{\tan^{- 1}y}$= $e^{2\tan^{- 1}y}$+ K
• D. $xe^{\tan^{- 1}y}$= tan^–1 y + K

Answer 1

Answer: (C)

$2xe^{\tan^{- 1}y}$= $e^{2\tan^{- 1}y}$+ K

Answer 2

Try to get $\frac{dx}{dy}$+ Px = Q