Question

Solution of $e^{\frac{dy}{dx}}=x$ when $x=1$ and $y=0$ is:

• A. $y=x(\log x-1)+4$
• B. $y=x(\log x-1)+3$
• C. $y=x(\log x+1)+1$
• D. $y=x(\log x-1)+1$

Answer 1

The correct option is D $y=x(\log x-1)+1$

$e^{\frac{dy}{dx}}=x$

$\Rightarrow \frac{dy}{dx}=\log x$

$\Rightarrow dy=\log xdx$

Let $\log x=t\Rightarrow x=e^t$ and $dx=e^{t}dt$

$\therefore dy=t{e}^{t}dt$

Integrating both sides,

$y=t{e}^t-\frac{dt}{dt}\int e^{t}dt+c$, where $c$ is a constant.

$\therefore y=t{e}^t-e^t+c$

i.e. $y=x(\log x-1)+c$

Substituting $y=0$ when $x=1$, we get

$0=1(\log 1-1)+c$

$\therefore c=1$

$\Rightarrow y=x(\log x-1)+1$