Solution of the differential equation [y(1+x-1)+ sin y] d x+(x+ log x+x cos y) d y=0 is

Question

Solution of the differential equation [y(1+x-1)+ sin y] d x+(x+ log x+x cos y) d y=0 is (A) x y+y log x=c (B) x y+x sin y=c (C) x y+y log x+x sin y=c (D) None of these

Tardigrade Admin · Accepted Answer

The given diff. equation can be written as ⇒ (y d x+x d y)+((y/x) d x+ log x) d y + sin y dx + x cos y dy = 0 ⇒ d(x y)+d(y log x)+d(x sin y)=0 Integrating, we get x y+y log x+x sin y=c.

Q. Solution of the differential equation
$[y (1 + x^{- 1}) + sin y] d x + (x + lo g x + x cos y) d y = 0$ is

$x y + y lo g x = c$

$x y + x sin y = c$

$x y + y lo g x + x sin y = c$

None of these

The given diff. equation can be written as
$\Rightarrow (y d x + x d y) + (\frac{y}{x} d x + lo g x) d y + sin y d x + x cos y d y = 0$
$\Rightarrow d (x y) + d (y lo g x) + d (x sin y) = 0$
Integrating, we get
$x y + y lo g x + x sin y = c$ .

Q. Solution of the differential equation [y(1+x−1)+siny]dx+(x+logx+xcosy)dy=0 is

xy+ylogx=c

xy+xsiny=c

xy+ylogx+xsiny=c