Question

x-y-2)dx-(2x-2y-3)dy=0 ​

Answer 1

Answer:

x + log (x - y - 1) = c₁ is the required answer.

Step-by-step explanation:

Given that (x - y - 2)dx -(2x - 2y - 3)dy = 0

=> (x - y - 2)dx = (2x - 2y - 3)dy

=> $\frac{dy}{dx}$ = $\frac{x-y-2}{2x-2y-3}$

=> Let (x - y - 2) = t

=> $\frac{dt}{dx}$ = 1 -  $\frac{dy}{dx}$ - 0 = 1 -  $\frac{dy}{dx}$

On substituting the value in above

=> 1 -  $\frac{dt}{dx}$  = $\frac{t}{2t+1}$

=>  $\frac{dt}{dx}$ = 1 - $\frac{t}{2t+1}$  = $\frac{2t+1 -t}{2t+1}$ = $\frac{t+1}{2t+1}$

Integrating both sides,

$\int {\frac{2t+1}{t+1} } \, dt$ = $\int\, dx$

=> $\int ({\frac{t}{t+1} + 1 }) \, dt$ = x + c

=>  $\int ({\frac{t}{t+1} ) \, dt$ + t = x + c                       --(i)

Now for solving   $\int ({\frac{t}{t+1} ) \, dt$

Let t + 1 = u

=> $\frac{du}{dt}$ = 1 + 0 = 1

=>  $\int ({\frac{t}{t+1} ) \, dt$  = $\int{\frac{u-1}{u}} \, du$

                     =  $\int{(1-\frac{1}{u}}) \, du$        (as $\int \, dx$ = x)

                     = u - log u             (as $\int\frac{1} {x} \, dx$ = log x)

Substituting this in equation (i)

=> u - log u + t = x + c

As u = t + 1

=> t + 1 - log(t + 1) + t = x + c

=> 1 - log (t + 1) = x + c

As x - y - 2 = t

=> 1 - log (x - y - 2 + 1) = x + c

=> 1 - log (x - y - 1) = x + c

=> 1 - c = x + log (x - y - 1)

=> x + log (x - y - 1) = c₁