The general solution of the exact differential equation $\left( \frac { y } { x - 1 } \right) d x + \left( \ln ( 2 x - 2 ) + \frac { 1 } { y } \right) d y = 0$ is

A) $- y \ln ( 2 x - 2 ) + \ln y = C$
B) $y \ln ( 2 x + 2 ) - \ln y = C$
C) $y \ln ( 2 x - 2 ) - \ln y = C$
D) $y \ln ( 2 x - 2 ) + \ln y = C$
E) $y \ln ( 2 x + 2 ) + \ln y = C$

Question

The general solution of the exact differential equation $\left( \frac { y } { x - 1 } \right) d x + \left( \ln ( 2 x - 2 ) + \frac { 1 } { y } \right) d y = 0$ is

A) $- y \ln ( 2 x - 2 ) + \ln y = C$
B) $y \ln ( 2 x + 2 ) - \ln y = C$
C) $y \ln ( 2 x - 2 ) - \ln y = C$
D) $y \ln ( 2 x - 2 ) + \ln y = C$
E) $y \ln ( 2 x + 2 ) + \ln y = C$

Quizplus · Accepted Answer

The Answer of The general solution of the exact differential equation $\left( \frac { y } { x - 1 } \right) d x + \left( \ln ( 2 x - 2 ) + \frac { 1 } { y } \right) d y = 0$ is

A) $- y \ln ( 2 x - 2 ) + \ln y = C$
B) $y \ln ( 2 x + 2 ) - \ln y = C$
C) $y \ln ( 2 x - 2 ) - \ln y = C$
D) $y \ln ( 2 x - 2 ) + \ln y = C$
E) $y \ln ( 2 x + 2 ) + \ln y = C$

The General Solution of the Exact Differential Equation (yx−1)dx+(ln⁡(2x−2)+1y)dy=0\left( \frac { y } { x - 1 } \right) d x + \left( \ln ( 2 x - 2 ) + \frac { 1 } { y } \right) d y = 0(x−1y​)dx+(ln(2x−2)+y1​)dy=0

The General Solution of the Exact Differential Equation $\left( \frac { y } { x - 1 } \right) d x + \left( \ln ( 2 x - 2 ) + \frac { 1 } { y } \right) d y = 0$