Question

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

• A. $y=k$
• B. $y=k x$
• C. $y=k\left(e^{x}+1\right)$
• D. $y=k\left(e^{x}-1\right)$

Answer 1

Answer: (D)

$\frac{y_{2}}{y_{1}}=1 \Rightarrow d\left(\log y_{1}\right) =1 $
$\Rightarrow \log 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 $y=k\left(e^{x}-1\right)$.