The differential -equation of the family of lines passing through the origin is

Question

The differential -equation of the family of lines passing through the origin is (A) x = y (dy/dx)  (B) x + y (dy/dx) = 0  (C)  (dy/dx) = 0  (D) y = x (dy/dx)

Tardigrade Admin · Accepted Answer

Equation of line passing. through-the origin is y = cx .........(i) Differentiating (i) w.r.t, x, we get (dy/dx) = c : : ⇒ (dy/dx) = (y/x) (From(i)) ⇒ : y = x (dy/dx)

Q. The differential -equation of the family of lines passing through the origin is

$x = y \frac{d y}{d x}$

$x + y \frac{d y}{d x} = 0$

$\frac{d y}{d x} = 0$

$y = x \frac{d y}{d x}$

Equation of line passing. through-the origin is $y = c x$ .........(i)
Differentiating (i) w.r.t, x, we get
$\frac{d y}{d x} = c \Rightarrow \frac{d y}{d x} = \frac{y}{x}$ (From(i))
$\Rightarrow y = x \frac{d y}{d x}$

Q. The differential -equation of the family of lines passing through the origin is

x=ydxdy​

x+ydxdy​=0

dxdy​=0

y=xdxdy​

Equation of line passing. through-the origin is y=cx .........(i) Differentiating (i) w.r.t, x, we get dxdy​=c⇒dxdy​=xy​ (From(i)) ⇒y=xdxdy​

$x = y \frac{d y}{d x}$

$x + y \frac{d y}{d x} = 0$

$\frac{d y}{d x} = 0$

$y = x \frac{d y}{d x}$

Equation of line passing. through-the origin is $y = c x$ .........(i)
Differentiating (i) w.r.t, x, we get
$\frac{d y}{d x} = c \Rightarrow \frac{d y}{d x} = \frac{y}{x}$ (From(i))
$\Rightarrow y = x \frac{d y}{d x}$