The family passing through (0,0) and satisfying the differential equation (y2/y1)=1 (where yn=(dn y/d xn) ) is

Question

The family passing through (0,0) and satisfying the differential equation (y2/y1)=1 (where yn=(dn y/d xn) ) is (A) y=k (B) y=k x (C) y=k(ex+1) (D) y=k(ex-1)

Tardigrade Admin · Accepted Answer

(y2/y1)=1 ⇒ d( log y1) =1 ⇒ log y1=x+c ⇒ y1=k ex ⇒ y=k ex+B Since it passes through (0,0), k+B=0 Thus, family is y=k(ex-1).

Q. The family passing through $(0, 0)$ and satisfying the differential equation $\frac{y _{2}}{y _{1}} = 1$ (where $y_{n} = \frac{d ^{n} y}{d x ^{n}}$ ) is

$y = k$

$y = k x$

$y = k (e^{x} + 1)$

$y = k (e^{x} - 1)$

$\frac{y _{2}}{y _{1}} = 1 \Rightarrow d (lo g y_{1}) = 1$
$\Rightarrow lo g y_{1} = x + c$
$\Rightarrow y_{1} = k e^{x}$
$\Rightarrow y = k e^{x} + B$
Since it passes through $(0, 0), k + B = 0$
Thus, family is $y = k (e^{x} - 1)$ .

Q. The family passing through (0,0) and satisfying the differential equation y1​y2​​=1 (where yn​=dxndny​ ) is

y=k

y=kx

y=k(ex+1)

y=k(ex−1)